A single serving of snack has 180 Calories (kilocalories). How many Joules of energy are in 1 serving of the snack? ( 1 cal = 4. 184J)

Answers

Answer 1

There are 753.12 Joules of energy in one serving of the snack. This means that when we eat this snack, our body will be able to use 753.12 joules of energy from the food.

Given, Calories = 180 Cal( 1 cal = 4. 184J)

We know, 1 calorie (cal) is equivalent to 4.184 Joules (J)

1 Calorie = 4.184 Joules (J)

Thus, 180 Cal (calories) = 180 × 4.184 J = 753.12 J

To find the number of joules of energy in one serving of the snack, we need to convert the given calories to joules because calories and joules are different units of energy. We use the following conversion factor: 1 calorie (cal) = 4.184 joules (J).

Therefore, we have to multiply the given calorie value by 4.184 to get the equivalent amount in joules. In this case, we are given that a single serving of the snack contains 180 calories.

To find the energy in joules, we use the formula:

E(J) = n(cal) x 4.184 (where E is energy in joules, n is the number of calories and 4.184 is the conversion factor).

Substituting the given values, we have:

E(J) = 180(cal) x 4.184

= 753.12 J

So, one serving of the snack has an energy of 753.12 joules (J).

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Related Questions

Tell whether the rates are equivalent.



0. 75 kilometer for every 30 minutes


1. 25 kilometers for every 50 minutes

Answers

No, the rates are not equivalent. Simplifying the first rate, we can say that 1 kilometer is covered in every 40 minutes. In the second rate, we can say that 1 kilometer is covered in every 2 minutes.

To determine if two rates are equivalent, we need to simplify the rates and compare the time it takes to cover one unit of distance. In the first rate, 0.75 kilometers are covered in 30 minutes. To simplify, we can divide both the numerator and denominator by 0.75, resulting in 1 kilometer covered in 40 minutes.

In the second rate, 25 kilometers are covered in 50 minutes. Simplifying by dividing both numerator and denominator by 25, we get 1 kilometer covered in 2 minutes.

Comparing the simplified rates, we see that it takes 40 minutes to cover 1 kilometer in the first rate, while it only takes 2 minutes in the second rate. Since the time required to cover the same distance differs, the rates are not equivalent.

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Atoms are found to move from one lattice position to another at the rate of 5×10^5jumpss at 400c° when the activation energy for their movement is 30,000 cal/mol. calculate the jump rate at 750c°.

Answers

The jump rate at 750°C is approximately [tex]1.84×10^24 jumps/s[/tex].

To calculate the jump rate at 750°C, we can use the Arrhenius equation:

[tex]k = A * exp(-Ea/RT)[/tex]

where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.

We are given that at 400°C, the jump rate is 5×10^5 jumps/s and the activation energy is 30,000 cal/mol. We need to find the jump rate at 750°C.

First, we need to convert the activation energy from calories per mole to joules per mole:

Ea = 30,000 cal/mol * 4.184 J/cal = 125,520 J/mol

Next, we need to convert the temperatures to Kelvin:

T1 = 400°C + 273.15 = 673.15 K

T2 = 750°C + 273.15 = 1023.15 K

Now we can use the Arrhenius equation to find the new jump rate:

[tex]k2 = A * exp(-Ea/RT2)[/tex]

We can solve for A by using the jump rate at 400°C:

[tex]5×10^5 jumps/s = A * exp(-Ea/RT1)[/tex]

[tex]A = 5×10^5 jumps/s * exp(Ea/RT1) = 5×10^5 jumps/s * exp(125,520 J/mol / (8.314 J/(mol·K) * 673.15 K)) = 6.95×10^12[/tex]

Now we can plug in A and the new temperature into the Arrhenius equation:

[tex]k2 = 6.95×10^12 * exp(-125,520 J/mol / (8.314 J/(mol·K) * 1023.15 K)) = 1.84×10^24[/tex]

Therefore, the jump rate at 750°C is approximately 1.84×10^24 jumps/s.

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Decreased susceptibility to the HIV virus has been associated with ____________________________. a. Major histocompatibility proteins b. CD4 proteins c. CCR5 delta32 cell surface proteins d. bone morphogenic proteins

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Decreased susceptibility to the HIV virus has been associated with CCR5 delta32 cell surface proteins. These proteins play a crucial role in HIV infection, as they are the main co-receptor for the virus to enter and infect cells.

Individuals who carry a genetic mutation that results in the deletion of the CCR5 delta32 protein have been found to have a higher level of resistance to HIV infection. This is because the virus is unable to enter and infect cells that lack the CCR5 delta32 protein. Research into this genetic mutation has led to the development of novel HIV therapies, such as gene editing techniques, that aim to mimic the protective effects of the CCR5 delta32 mutation.


Decreased susceptibility to the HIV virus has been associated with CCR5 delta32 cell surface proteins. The CCR5 delta32 variant leads to a nonfunctional receptor, which inhibits the entry of HIV into cells. This genetic mutation provides individuals with some level of resistance to the virus, as it prevents the virus from binding to CD4 T cells, an essential step for infection. While major histocompatibility proteins, CD4 proteins, and bone morphogenic proteins play important roles in immune system function, they are not directly linked to decreased susceptibility to HIV as CCR5 delta32 cell surface proteins are.

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A group of students performed the aspirin experiment. They prepared a stock solution that was 0.008450 mol/L in ASA. Then they prepared a standard solution by transferring 4.97 mL of the stock solution to a 50-mL volumetric flask and diluting to the mark with FeCl3-KCl-HCl solution. What was the concentration of the standard solution in mol/L

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The concentration of the standard solution can be calculated using the principles of dilution. By transferring a known volume of the stock solution to a volumetric flask and diluting it to the mark, the concentration of the standard solution can be determined. In this case, the stock solution has a known concentration of 0.008450 mol/L, and 4.97 mL of the stock solution is transferred to a 50-mL volumetric flask.

To find the concentration of the standard solution, we use the formula for dilution:

C1V1 = C2V2

Where C1 is the concentration of the stock solution, V1 is the volume of the stock solution transferred, C2 is the concentration of the standard solution, and V2 is the final volume of the standard solution.

In this case, we have:

C1 = 0.008450 mol/L (concentration of the stock solution)

V1 = 4.97 mL (volume of the stock solution transferred)

C2 = ? (concentration of the standard solution)

V2 = 50 mL (final volume of the standard solution)

Substituting the given values into the dilution formula, we can solve for C2:

(0.008450 mol/L)(4.97 mL) = C2(50 mL)

C2 = (0.008450 mol/L)(4.97 mL) / (50 mL)

C2 ≈ 0.000839 mol/L

Therefore, the concentration of the standard solution is approximately 0.000839 mol/L.

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Consider the following system at equilibrium where Kc = 1.80×10-4 anddelta16-1.GIFH° = 92.7 kJ/mol at 298 K.NH4HS (s)Doublearrow.GIFNH3 (g) + H2S (g)The production of NH3 (g) is favored by:Indicate True (T) or False (F) for each of the following:___TF 1. increasing the temperature.___TF 2. decreasing the pressure (by changing the volume).___TF 3. increasing the volume.___TF 4. adding NH4HS .___TF 5. removing H2S .

Answers

Increasing the temperature (False), decreasing the pressure (True), increasing the volume (True), adding NH4HS (True), and removing H2S (True) favor the production of NH3 (g).

The production of NH3 (g) is favored by:


1. False - Increasing the temperature will not favor the production of NH3 (g) since it is an exothermic reaction (ΔH° = 92.7 kJ/mol).
2. True - Decreasing the pressure (by changing the volume) will favor the production of NH3 (g) as it increases the number of gas molecules on the right side of the reaction.
3. True - Increasing the volume will also favor the production of NH3 (g) as it shifts the equilibrium towards the side with more gas molecules (right side).
4. True - Adding NH4HS will favor the production of NH3 (g) as the equilibrium shifts to the right to counteract the increase in the reactant.
5. True - Removing H2S will favor the production of NH3 (g) as the equilibrium shifts to the right to replace the removed product.

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A 30.0-ml sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. Calculate the at 0 of added base.

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The pH of the 0.165 M propanoic acid solution at 0 mL of added 0.300 M KOH is 4.87.

To calculate the pH at the beginning of the titration (0 mL of added base), we'll use the information given about the propanoic acid solution.

The formula for calculating the pH of a weak acid is:
pH = pKa + log([A-]/[HA])

First, we need to find the pKa for propanoic acid. The Ka for propanoic acid is 1.34 x 10^-5. Using the formula pKa = -log(Ka), we find:

pKa = -log(1.34 x 10^-5) = 4.87

Since no base has been added, the ratio of [A-]/[HA] is 0, and the log term becomes 0 as well. So, the pH is equal to the pKa at this point:

pH = 4.87

Therefore, the pH of the 0.165 M propanoic acid solution at 0 mL of added 0.300 M KOH is 4.87.

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in an acid-base reaction where ha acts as an acid, what will be the conjugate base?

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The species that is created after a chemical like HA donates a proton (H⁺) acting as an acid in an acid-base reaction is known as the conjugate base.

A proton is taken out of the original acid to create the conjugate base. The overall response can be pictured as follows: Acid + Water + Conjugate Base + H₃O⁺. The acid that provides a proton (H⁺) is called HA.

The hydronium ion (H₃O⁺) is formed when the proton is taken up by the base H₂O. The conjugate base that results from HA losing a proton is called A.

The species that remains after an acid (HA) loses a proton and is capable of taking a proton to regenerate the initial acid (HA) is the conjugate base, A.

Thus, The species that is created after a chemical like HA donates a proton (H⁺) acting as an acid in an acid-base reaction is known as the conjugate base.

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find [OH-], [H+], and the pH and the pOH of the followingsolutions,a) 0.27 M Sr(OH)2b) a solution made by dissolving 13.6 g of KOH in enough water tomake 2.50 L of solution.

Answers

The pH and the pOH of the solutions is: a) For the 0.27 M Sr(OH)₂ solution, [OH⁻] is 0.54 M, [H⁺] is 1.85×10⁻¹² M, pH is 12.26 and pOH is 1.74. b) For the solution made by dissolving 13.6 g of KOH in enough water, [OH⁻] is 2.67 M, [H⁺] is 3.75×10⁻¹⁴ M, pH is 13.43 and pOH is 0.57.

a) Since Sr(OH)₂ dissociates in water to produce two moles of OH⁻ for every mole of Sr(OH)₂, the concentration of OH⁻ in the solution will be twice the concentration of Sr(OH)₂.

Therefore:

[OH⁻] = 2 × 0.27 M = 0.54 M

Using the expression for the ion product of water (Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C), we can calculate [H⁺]:

[H⁺] = Kw/[OH⁻] = (1.0×10⁻¹⁴)/(0.54) = 1.85×10⁻¹² M

Taking the negative logarithm of [H⁺] gives the pH:

pH = -log[H⁺] = -log(1.85×10⁻¹²) = 12.26

The pOH can be calculated as:

pOH = -log[OH⁻] = -log(0.54) = 1.74

b) The molar mass of KOH is 56.11 g/mol, so 13.6 g of KOH corresponds to 13.6/56.11 mol = 0.243 mol.

The concentration of KOH in the solution is therefore:

0.243 mol/2.50 L = 0.097 M

KOH is a strong base, so it completely dissociates in water to produce one mole of OH⁻ for every mole of KOH. Therefore:

[OH⁻] = 0.097 M

Using Kw, we can calculate [H⁺]:

[H⁺] = Kw/[OH⁻] = (1.0×10⁻¹⁴)/(0.097) = 3.75×10⁻¹⁴ M

Taking the negative logarithm of [H⁺] gives the pH:

pH = -log[H⁺] = -log(3.75×10⁻¹⁴) = 13.43

The pOH can be calculated as:

pOH = -log[OH⁻] = -log(0.097) = 0.57

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mno−4(aq) cr(oh)3(s)⟶cro2−4(aq) mno2(s) how many hydroxide ions will appear in the balanced equation?

Answers

The reaction of MnO4- and Cr(OH)3 to produce CrO42- and MnO2 has the following balanced equation:

3CrO42-(aq) + 2MnO2(s) + 6OH-(aq) = 2MnO4-(aq) + 3Cr(OH)3(s)

Six hydroxide ions (OH-) will show up on the reaction's product side, according to the balanced equation. This is due to the fact that each Cr(OH)3 molecule provides two hydroxide ions to the process, which requires three molecules of Cr(OH)3 to react with two molecules of MnO4-. As a result, the reaction produces a total of 6 hydroxide ions (2 x 3).

Thus, the balanced equation demonstrates that the reaction of 2MnO4-(aq) and 3Cr(OH)3(s) to form 3CrO42-(aq) and 2MnO2(s) results in the production of six hydroxide ions.

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The balanced equation demonstrates that the reaction of 2MnO4-(aq) and 3Cr(OH)3(s) to form 3CrO42-(aq) and 2MnO2(s) results in the production of six hydroxide ions.

The reaction of MnO4- and Cr(OH)3 to produce CrO42- and MnO2 has the following balanced equation:

3CrO42-(aq) + 2MnO2(s) + 6OH-(aq) = 2MnO4-(aq) + 3Cr(OH)3(s)

Six hydroxide ions (OH-) will show up on the reaction's product side, according to the balanced equation. This is due to the fact that each Cr(OH)3 molecule provides two hydroxide ions to the process, which requires three molecules of Cr(OH)3 to react with two molecules of MnO4-. As a result, the reaction produces a total of 6 hydroxide ions (2 x 3). Thus, the balanced equation demonstrates that the reaction of 2MnO4-(aq) and 3Cr(OH)3(s) to form 3CrO42-(aq) and 2MnO2(s) results in the production of six hydroxide ions.

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I am confused with how to do this. Could someone help me.
1. Write a balanced nuclear equation for the following:
a. bismuth-211 undergoes beta decay:
b. chromium-50 undergoes positron emission:
c. mercury-188 decays to gold-188:
d. plutonium-242 undergoes alpha emission:

Answers

Here are the balanced nuclear equations for each of the four given scenarios:

a. Bismuth-211 undergoes beta decay:
Bi-211 (83) -> Po-211 (84) + β^-

b. Chromium-50 undergoes positron emission:
Cr-50 (24) -> V-50 (23) + β^+

c. Mercury-188 decays to gold-188:
Hg-188 (80) -> Au-188 (79) + β^-

d. Plutonium-242 undergoes alpha emission:
Pu-242 (94) -> U-238 (92) + α

In each equation, the element symbol is accompanied by its mass number, and the atomic number is shown in parentheses.

The emitted particles are represented by their respective symbols (β^- for beta decay, β^+ for positron emission, and α for alpha emission).

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Predict which bond in each of the following sis the most polar
a.C-F, si-F, Ge-F
b. P-Cl, S-Cl
c. S-F, S-Cl, S-Br
d. Ti-Cl, Si-Cl, Ge-Cl

Answers

(a) Among C-F, Si-F, and Ge-F, the C-F bond is the most polar because fluorine (F) is more electronegative than carbon (C), silicon (Si), and germanium (Ge),

which results in a greater difference in electronegativity and a more polar bond.

(b) Among P-Cl and S-Cl, the S-Cl bond is the most polar because sulfur (S) is more electronegative than phosphorus (P),

which results in a greater difference in electronegativity and a more polar bond.

(c) Among S-F, S-Cl, and S-Br, the S-F bond is the most polar because fluorine (F) is the most electronegative element in this group,

resulting in the greatest difference in electronegativity and the most polar bond.

(d) Among Ti-Cl, Si-Cl, and Ge-Cl, the Si-Cl bond is the most polar because chlorine (Cl) is more electronegative than silicon (Si) and germanium (Ge),

But titanium (Ti) is more electronegative than both silicon and germanium, which results in a smaller difference in electronegativity and a less polar bond.

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A radiation source of 1000 watts is located at a point in space. What is the intensity of radiation at a distance of 10 meters form the source

Answers

The intensity of radiation from a source follows an inverse square law, which means that as the distance from the source increases, the intensity decreases.

Given:
Power of the radiation source = 1000 watts
Distance from the source = 10 meters

The intensity (I) of radiation is defined as the power (P) per unit area (A):

Intensity = Power / Area

Since we are not given the specific area, we need to make an assumption. Let's assume that the radiation is spreading out equally in all directions, forming a spherical wavefront.

The surface area of a sphere is given by the formula:
Area = 4πr^2

Where r is the distance from the source.

Plugging in the values:
Area = 4π(10)^2 = 400π square meters

Now we can calculate the intensity:
Intensity = Power / Area
Intensity = 1000 watts / 400π square meters

To round the answer to three significant figures, we can use 3.14 as an approximation for π.

Intensity ≈ 1000 watts / (400 * 3.14) square meters
Intensity ≈ 0.795 watts per square meter

Therefore, at a distance of 10 meters from the source, the intensity of radiation is approximately 0.795 watts per square meter.

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3. the concentration of a sodium hydroxide solution is to be determined. a 50.0 ml sample of 0.104 m hcl solution requires 48.7 ml of the sodium hydroxide solution to reach the point of neutralization

Answers

The concentration of the sodium hydroxide solution is approximately 0.106 M.

Titration is a technique that can be used to figure out how much sodium hydroxide (NaOH) is in a solution. Titration is a method that determines the concentration of an unknown solution, in this case, the 0.104 M HCl solution, using a solution of known concentration.

Use the following formula to accomplish this:

M1V1 = M2V2

Where M1 and V1 stand for the HCl solution's volume and molarity, respectively, and M2 and V2 for the NaOH solution's, respectively. Our information includes the following:

M1 (HCl) = 0.104 M
V1 (HCl) = 50.0 mL
V2 (NaOH) = 48.7 mL

We need to find M2, which is the concentration of the NaOH solution. Plugging the given values into the formula, we have:

(0.104 M)(50.0 mL) = (M2)(48.7 mL)

Now, we can solve for M2:

M2 = (0.104 M)(50.0 mL) / (48.7 mL)

M2 ≈ 0.106 M

So, the concentration of the sodium hydroxide solution is approximately 0.106 M.

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how many rings are present in c18h29bro3? this compound consumes 2 mol of h2 on catalytic hydrogenation. enter your answer in the provided box.

Answers

In the compound C18H29BrO3, there are 7 rings present. However, we don't have enough information about the connectivity of the atoms in the molecule. Therefore, it is not possible to give a detailed answer to this question without additional information.

Regarding the second part of the question, catalytic hydrogenation of c18h29bro3 consumes 2 mol of h2, which means that each molecule of the compound reacts with two molecules of hydrogen gas. This information can be used to calculate the stoichiometry of the reaction and the amount of product formed under specific conditions.

When the compound consumes 2 moles of H2 during catalytic hydrogenation, it means that two double bonds or other unsaturated bonds are present. The general formula for an acyclic alkane is CnH(2n+2). Since this compound has 18 carbons, the number of hydrogens in a saturated alkane would be 2(18) + 2 = 38.

Now, let's compare the actual number of hydrogens in the given compound with the expected number for a saturated alkane. The compound has 29 hydrogens, which is 9 less than the expected number (38 - 29 = 9).

Considering that it consumed 2 moles of H2, we can infer that there are 2 double bonds or other unsaturated bonds (each consuming 1 mole of H2) in the compound. This means there are 7 remaining unsaturations that can be attributed to rings. So, in the compound C18H29BrO3, there are 7 rings present.

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A buffer is prepared by adding 12.0 grams of ammonium chloride (NH4Cl) to 260 mL of 1.00 M NH33 solution.
a. What is the pH of this buffer?
b. Write the net ionic equation for the reaction that occurs when a few drops of nitric acid is added to the buffer.
c. Write the net ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution is added to the buffer.

Answers

NH4+ (aq) + OH- (aq) → NH3 (aq) + H2O (l)  

a. To find the pH of the buffer, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid])

The pKa of ammonium chloride is 9.25. Ammonium chloride acts as an acid in water, and ammonia acts as a base. Therefore, NH4+ is the acid and NH3 is the base.

First, we need to find the concentration of NH4+ and NH3 in the buffer:

moles NH4Cl = 12.0 g / 53.49 g/mol = 0.224 mol NH4Cl
moles NH3 = 260 mL x 1.00 M = 0.260 mol NH3

Since NH4Cl dissociates completely in water, all the NH4+ in the solution comes from the NH4Cl added. Therefore, the concentration of NH4+ is 0.224 mol / 0.260 L = 0.862 M.

The concentration of NH3 is already given as 1.00 M.

Now we can plug these values into the Henderson-Hasselbalch equation:

pH = 9.25 + log(1.00 / 0.862) = 9.02

Therefore, the pH of the buffer is 9.02.

b. When a few drops of nitric acid is added to the buffer, it will react with the NH3 base to form ammonium nitrate, NH4NO3:

NH3 + HNO3 → NH4NO3

The net ionic equation for this reaction is:

NH3 + H+ → NH4+

c. When a few drops of potassium hydroxide solution is added to the buffer, it will react with the NH4+ acid to form ammonia and water:

NH4+ + OH- → NH3 + H2O

The net ionic equation for this reaction is:

H+ + OH- → H2O (this is the neutralization reaction)
a. To find the pH of the buffer, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

First, we need to calculate the concentration of NH4Cl and NH3 in the buffer solution. The molar mass of NH4Cl is 53.49 g/mol.

12.0 g NH4Cl * (1 mol NH4Cl / 53.49 g NH4Cl) = 0.224 mol NH4Cl

The volume of the solution is 0.260 L. Therefore, the concentration of NH4Cl (A-) is:

0.224 mol NH4Cl / 0.260 L = 0.862 M

The concentration of NH3 (HA) is given as 1.00 M. The pKa of NH4+ is 9.25. Now we can plug these values into the Henderson-Hasselbalch equation:

pH = 9.25 + log (0.862 / 1.00) = 9.25 - 0.064 = 9.19

The pH of the buffer is 9.19.

b. The net ionic equation for the reaction when a few drops of nitric acid (HNO3) are added to the buffer is:

NH3 (aq) + H+ (aq) → NH4+ (aq)

c. The net ionic equation for the reaction when a few drops of potassium hydroxide (KOH) solution are added to the buffer is:

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the temperature of a sample of silver increased by 23.1 °c when 255 j of heat was applied. what is the mass of the sample?
_____g
substance specific heat j/(g*c)
lead 0.128
silver 0.235
copper 0.385
iron 0.449
aluminum 0.903

Answers

To find the mass of the sample of silver, we can use the formula: q = mcΔT. Where q is the amount of heat energy absorbed, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Plugging in the values we have:

255 J = m x 0.235 J/(g°C) x 23.1°C

Simplifying, we get:

255 J = 5.4335 m

Dividing both sides by 5.4335, we get:

m = 46.9 g

Therefore, the mass of the sample of silver is 46.9 g.


To find the mass of the silver sample when the temperature increased by 23.1°C and 255 J of heat was applied, you can use the formula:

Q = mcΔT

where Q is the heat energy (255 J), m is the mass of the sample (in grams), c is the specific heat capacity of the substance (in J/(g°C)), and ΔT is the temperature change (23.1°C).

For silver, the specific heat capacity is 0.235 J/(g°C). Now we can rearrange the formula to solve for the mass (m):

m = Q / (cΔT)

Plugging in the given values:

m = 255 J / (0.235 J/(g°C) × 23.1°C)

m ≈ 47.45 g

The mass of the sample is approximately 47.45 grams.

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What is the freezing point of a solution that contains 12.0 g of glucose in 50 g of acetic acid (CH3COOH). Kf = 3.90°C/m, melting point = 16.6 °C, [C6H12O6 – MM 180.2 g/mol].

Answers

To determine the freezing point of the solution, we need to use the formula: ΔTf = Kf × molality. Where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent (acetic acid), and molality is the concentration of the solute (glucose) in moles per kilogram of solvent.

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

The molar mass of glucose (C6H12O6) is 180.2 g/mol, so we have:

moles of glucose = 12.0 g / 180.2 g/mol = 0.0665 mol

mass of acetic acid = 50 g / 1000 g/kg = 0.05 kg

molality = 0.0665 mol / 0.05 kg = 1.33 mol/kg

Now we can plug in the values for Kf and molality to find ΔTf:

ΔTf = 3.90°C/m × 1.33 mol/kg = 5.19°C

Finally, we can calculate the freezing point of the solution:

freezing point = melting point - ΔTf

freezing point = 16.6°C - 5.19°C = 11.41°C

Therefore, the freezing point of the solution is 11.41°C.

To find the freezing point of a solution containing 12.0 g of glucose in 50 g of acetic acid, we can use the formula ΔTf = Kf × molality. First, calculate the molality by dividing moles of glucose by the mass of acetic acid in kilograms:

Moles of glucose = 12.0 g / 180.2 g/mol = 0.0666 mol
Mass of acetic acid = 50 g / 1000 = 0.05 kg

Molality = 0.0666 mol / 0.05 kg = 1.332 mol/kg

Now, calculate ΔTf:

ΔTf = Kf × molality = 3.90°C/m × 1.332 mol/kg = 5.1948 °C

Finally, subtract ΔTf from the melting point of acetic acid:

Freezing point of the solution = 16.6 °C - 5.1948 °C = 11.4052 °C

The freezing point of the solution is approximately 11.41 °C.

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Here is a table of densities of common metals. Use the table to identify the metal in each case: Name of metal Density (g/cm^3) magnesium 1.74 aluminum 2.72titanium 4.5vanadium 5.494zinc 7.14 steel 7.85 brass 8.52 copper 10.5silver 8.94 lead 11.3 palladium 12.0gold 19.3platinum 21.4

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The provided table lists the densities of various common metals. By comparing the given densities, we can identify the corresponding metals, such as magnesium, aluminum, titanium, vanadium, zinc, steel, brass, copper, silver, lead, palladium, gold, and platinum.

Based on the provided table, we can identify the metals as follows:

1. The metal with a density of 1.74 g/cm³ is magnesium.

2. The metal with a density of 2.72 g/cm³ is aluminum.

3. The metal with a density of 4.5 g/cm^³ is titanium.

4. The metal with a density of 5.494 g/cm³ is vanadium.

5. The metal with a density of 7.14 g/cm³ is zinc.

6. The metal with a density of 7.85 g/cm³ is steel.

7. The metal with a density of 8.52 g/cm³ is brass.

8. The metal with a density of 10.5 g/cm³ is copper.

9. The metal with a density of 8.94 g/cm³ is silver.

10. The metal with a density of 11.3 g/cm³ is lead.

11. The metal with a density of 12.0 g/cm³ is palladium.

12. The metal with a density of 19.3 g/cm³ is gold.

13. The metal with a density of 21.4 g/cm³ is platinum.

By matching the densities with the corresponding metals, we can identify the specific metal in each case.

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Given the values of ΔH∘rxn, ΔS∘rxn, and T below, determine ΔSuniv. δh∘rxn=− 132 kj , δs∘rxn=− 259 j/k , t= 564 k .

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The value of ΔSuniv is the change in the universe's entropy, which measures how chaotic or unpredictable a process is as it happens during a chemical or physical reaction. Thus, ΔSuniv = 0 J/K.

To determine ΔSuniv, we use the equation ΔSuniv = ΔSsys + ΔSsurr, where ΔSsys is the change in entropy of the system and ΔSsurr is the change in entropy of the surroundings. We can calculate ΔSsys using the equation ΔSsys = ΔH∘rxn / T, where T is the temperature in Kelvin.
ΔSsys = (-132 kJ) / (564 K) = -0.234 J/K
To calculate ΔSsurr, we use the equation ΔSsurr = -ΔH∘rxn / T. This is because the surroundings will have an opposite change in entropy to that of the system.
ΔSsurr = -(-132 kJ) / (564 K) = 0.234 J/K
Now we can calculate ΔSuniv by adding ΔSsys and ΔSsurr.
ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv = -0.234 J/K + 0.234 J/K
ΔSuniv = 0 J/K
Therefore, the value of ΔSuniv is 0 J/K.

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Briefly explain the meanings of the following terms as they relate to this experiment. Include structural formulas if appropriate. (1) aldohexose (2) reducing sugar (3) hemiacetal

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Aldohexose is a six-carbon sugar that contains an aldehyde group. A reducing sugar is a sugar that has a free aldehyde or ketone group, and a hemiacetal is a functional group that results from the reaction of an aldehyde with an alcohol.

What is the meaning of aldohexose, reducing sugar, and hemiacetal in the context of the experiment?

(1)Aldohexose: It is a type of monosaccharide or simple sugar that contains six carbon atoms and an aldehyde functional group (-CHO) on the first carbon atom.

Glucose, the most common aldohexose is an important source of energy for living organisms.

(2)Reducing sugar: It is a type of sugar that has the ability to reduce certain chemicals by donating electrons. In the context of this experiment, a reducing sugar is a sugar that can react with Benedict's reagent, resulting in the formation of a colored precipitate.

Examples of reducing sugars include glucose, fructose, maltose, and lactose.

(3)Hemiacetal: It is a functional group that forms when an aldehyde or ketone reacts with an alcohol. In the context of this experiment, the reaction between the aldehyde group of a reducing sugar and an alcohol group of another molecule leads to the formation of a hemiacetal. This reaction is important in the Benedict's test for reducing sugars.

The hemiacetal formation between the reducing sugar and copper ions from the Benedict's reagent leads to the formation of a colored precipitate.

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which species has the strongest carbon - carbon bond, c2hcl , c2h6 , or c2cl4 ?

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The species with the strongest carbon-carbon bond is C₂H₆ (ethane). Ethane consists of two carbon atoms that are bonded together by a single sigma bond, which is the strongest type of covalent bond.

When two atoms form a covalent bond, they share a pair of electrons to achieve a stable electron configuration. In the case of multiple bonds between carbon atoms, there is a higher electron density and longer bond length compared to single bonds.

This is because the additional bonds share more electrons and have a larger electron cloud, leading to a weaker bond.  The introduction of electronegative atoms such as chlorine into a molecule can also affect the strength of carbon-carbon bonds. Chlorine has a higher electronegativity than carbon, meaning it attracts electrons more strongly.

As a result, the electrons in the bond are pulled towards the chlorine atom, creating partial charges and making the bond less symmetrical. This reduces the overlap of the electron clouds of the carbon atoms, leading to a weaker bond.

Ethane, on the other hand, has a simple single bond between its two carbon atoms, where the electrons are evenly shared. This results in a more symmetrical bond and stronger overlap of the electron clouds, leading to a stronger carbon-carbon bond.

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true/false. the avr uses the term twi instead of i2c.

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True.

AVR, which stands for Advanced Virtual RISC, uses the term TWI (Two-Wire Interface) instead of I2C (Inter-Integrated Circuit) to refer to a communication protocol that allows for simple, two-wire serial communication between multiple devices on a shared bus.

TWI and I2C are very similar protocols, but TWI is specific to AVR microcontrollers, while I2C is a more general protocol used by many different manufacturers.

The TWI protocol was developed by Atmel (now part of Microchip Technology) specifically for their AVR microcontrollers, and it is essentially a subset of the I2C protocol. So while the two protocols are very similar, they are not exactly the same.

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How many coulombs of charge are required to cause reduction of .25 moles of Cu2+ to Cu?A) .25 CB) .30 CC) 1.2 x 10^4D) 2.4 x 10^4E) 4.8 x 10^4

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Ok, let's break this down step-by-step:

* 0.25 moles of Cu2+ ions

* Each Cu2+ ion has a charge of +2

* So 0.25 moles of Cu2+ ions = 0.25 * 2 = 0.5 moles of positive charge

* To reduce Cu2+ to Cu, we need to provide an equal amount of negative charge (electrons)

* 1 mole of electrons = 1 faraday = 96485 C

* So 0.5 moles of electrons needed = 0.5 * 96485 C

* 0.5 * 96485 C = 47425 C

Therefore, the answer is B: 0.30 coulombs (round 47425 C to the nearest choice)

The required coulombs of charge for the reduction of 0.25 moles of Cu2+ to Cu is 48,242.5 C, which is approximately equal to 4.8 x 10⁴ C. Therefore, the correct answer is E) 4.8 x 10⁴.

To determine the number of coulombs required to cause the reduction of 0.25 moles of Cu2+ to Cu, we need to consider the balanced redox reaction and Faraday's constant. Here's the step-by-step explanation:

Step 1: Write the balanced redox reaction for the reduction of Cu2+ to Cu:
Cu2+ + 2e- → Cu

Step 2: Calculate the number of moles of electrons (e-) required for the reaction:
Since 1 mole of Cu2+ requires 2 moles of e-, 0.25 moles of Cu2+ will require 0.25 * 2 = 0.5 moles of e-.

Step 3: Convert the moles of electrons to coulombs using Faraday's constant (1 mole of e- = 96,485 C):
0.5 moles of e- * 96,485 C/mole = 48,242.5 C

The required coulombs of charge for the reduction of 0.25 moles of Cu2+ to Cu is 48,242.5 C, which is approximately equal to 4.8 x 10⁴ C. Therefore, the correct answer is E) 4.8 x 10⁴.

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calculate the temp. (in kelvin) of a 1.50 mol of a sample of a gas 1.25 atm and a volume of 14 L

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The temperature (in kelvin) of a 1.50 mol of a sample of a gas 1.25 atm and a volume of 14 L is 142.1 K

The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as

                                 PV = nRT

where,

P = Pressure

V = Volume

T = Temperature

n = number of moles

Given,

number of moles = 1.5 moles

pressure = 1.25 atm

volume = 14 L

PV = nRT

1.25 × 14 = 1.5 × 0.0821 × T

T = 142.1 K

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how can insulating zro2 be made into an electronic conductor

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Insulating zirconia ( [tex]ZrO_2[/tex]) can be made into an electronic conductor by introducing dopants, which are atoms or molecules that are added to the material to change its properties.

These dopants can create oxygen vacancies in the [tex]ZrO_2[/tex] lattice, which can then act as electron carriers and enable the material to conduct electricity. Some common dopants used for zirconia include yttria (Y2O3), ceria (CeO2), and alumina ([tex]Al_2O_3[/tex]). By carefully controlling the dopant concentration and processing conditions, it is possible to tailor the electronic properties of [tex]ZrO_2[/tex] to meet specific application requirements, such as in fuel cells, sensors, and electronic devices.

In summary, insulating  [tex]ZrO_2[/tex] can be made into an electronic conductor by doping it with impurities like [tex]Y_2O_3[/tex] or CaO, which create oxygen vacancies and ionic conductivity, leading to electronic conductivity in the material.

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calculate the wavelength (in m) of a football (425 g) thrown by an nfl quarterback traveling at 50 mph.

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The wavelength of the football thrown by an NFL quarterback traveling at 50 mph is approximately 6.99 x 10^-35 m.

To calculate the wavelength of the football, we need to first calculate its velocity in meters per second.

We can convert 50 mph to meters per second as follows:

1 mph = 0.44704 m/s (conversion factor)

50 mph = 50 x 0.44704 m/s

50 mph = 22.352 m/s (velocity of the football)

Next, we need to calculate the momentum of the football using the equation:

p = mv , where p is momentum, m is mass, and v is velocity.

We can convert the mass of the football from grams to kilograms as follows:

425 g = 0.425 kg (conversion factor)

So, the momentum of the football is:

p = mv

p = 0.425 kg x 22.352 m/s

p = 9.498 kg*m/s

Finally, we can calculate the wavelength of the football using the equation:

wavelength = h/p

where h is Planck's constant (6.626 x 10^-34 J*s).

So, the wavelength of the football is:

wavelength = h/p

wavelength = (6.626 x 10^-34 Js)/(9.498 kgm/s)

wavelength = 6.99 x 10^-35 m

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The wavelength of the football is λ = 7.17 * 10^-{26} nm .

The wavelength of the football can be calculated using the de Broglie wavelength equation: λ = h/mv, where h is Planck's constant, m is the mass of the object, v is the velocity of the object.
First, we need to convert the mass of the football from grams to kilograms: 425 g = 0.425 kg.
Next, we need to convert the velocity from mph to m/s: 50 mph = 22.35 m/s.
Now we can plug in the values into the equation:
λ = \frac{(6.626 * 10^{-34} J*s) }{ (0.425 kg * 22.35 m/s) }
λ = 7.17 * 10^{-26} nm
Therefore, the correct answer is C) 7.17 * 10^-{26} nm.
It's important to note that this calculation assumes that the football is behaving as a wave, which is not necessarily the case in reality. However, this calculation can still provide a useful estimate of the football's wavelength.

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How is the AHf related to the AH of a reaction?

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The enthalpy of formation (ΔHf) is related to the enthalpy change (ΔH) of a reaction through Hess's law, which states that the enthalpy change of a reaction can be calculated by the difference in enthalpies of formation of the products and reactants.

Enthalpy of formation (ΔHf) refers to the enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states at a given temperature and pressure. It is typically measured in kilojoules per mole (kJ/mol).

Hess's law states that the enthalpy change of a reaction is equal to the difference in enthalpies of formation between the products and reactants. In other words, if the enthalpies of formation of the products and reactants are known, the enthalpy change of the reaction can be calculated by taking the difference between them.

Mathematically, it can be represented as:

ΔH = Σ(nΔHf products) - Σ(nΔHf reactants)

Where ΔH is the enthalpy change of the reaction, n represents the stoichiometric coefficients of the compounds involved, and ΔHf is the enthalpy of formation.

Therefore, the enthalpy of formation (ΔHf) is a key component in calculating the enthalpy change (ΔH) of a reaction using Hess's law, as it provides the necessary values for the reactants and products involved in the reaction.

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FILL IN THE BLANK the reaction of 50 ml of cl2 gas with 50 ml of ch4 gas via the equation: cl2(g) ch4(g)→hcl(g) ch3cl(g) will produce a total of __________ ml of products if pressure and temperature are kept constant.

Answers

The reaction of 50 mL of Cl₂ gas with 50 mL of CH₄ gas via the equation: Cl₂(g) + CH₄(g) → HCl(g) + CH₃Cl(g) will produce a total of 100 mL of products if pressure and temperature are kept constant.

According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

In this reaction, one mole of Cl₂ reacts with one mole of CH₄ to produce one mole of HCl and one mole of CH₃Cl. Since the volumes of reactants are equal (50 mL each), and the mole ratio is 1:1 for both reactants and products, the total volume of products formed will be the sum of the individual volumes of the reactants, which is 50 mL + 50 mL = 100 mL. This holds true as long as the pressure and temperature conditions remain constant throughout the reaction.

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Refer to the information above. If you have 100g of calcium phosphate (310. 18g) and an excess of silicon dioxide and carbon, how many moles of phosphorus(30. 97) will be produced?

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When 100g of calcium phosphate (Ca3(PO4)2) reacts with an excess of silicon dioxide (SiO2) and carbon (C), the amount of phosphorus (P) produced can be calculated. The molar mass of calcium phosphate is 310.18g/mol, and the molar mass of phosphorus is 30.97g/mol.

Number of moles of calcium phosphate = 100g / 310.18g/mol

Next, we can use the balanced chemical equation to determine the stoichiometric ratio between calcium phosphate and phosphorus. From the equation, we can see that one mole of calcium phosphate produces one mole of phosphorus:

Number of moles of phosphorus = Number of moles of calcium phosphate

Therefore, the number of moles of phosphorus produced will be equal to the number of moles of calcium phosphate, which can be calculated using the given mass and molar mass of calcium phosphate.

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how many chirality centers are present in trans cinnamic acid? does cinnamic acid exist in any stereoisomeric form? if so how many stereoisomers are expected for cinnamic acid?

Answers

Trans-cinnamic acid has one chirality center, which is the carbon atom that is directly attached to the carboxylic acid group (-COOH). This carbon atom is sp² hybridized and has three different groups attached to it: a hydrogen atom, a double bond with an adjacent carbon, and a carboxylic acid group.

Due to this, two stereoisomers are possible for trans-cinnamic acid: (E)-cinnamic acid and (Z)-cinnamic acid. The (E)-isomer has the two highest priority groups (i.e., the double bond and the carboxylic acid group) on opposite sides of the double bond, whereas the (Z)-isomer has them on the same side of the double bond.

Both isomers have the same chirality center, but they differ in their geometric arrangement around the double bond. Therefore, cinnamic acid exists in two stereoisomeric forms, (E)-cinnamic acid and (Z)-cinnamic acid.

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