Answers
Answer:
2Ch3cooh + mg(oh)2 --> (ch3coo)2mg + 2h20
Explanation:
Reaction of milk of magnesia with acetic acid.
A solution of milk of magnesia and universal indicator in water is stirred while acetic acid is added. The indicator changes color as the acid is neutralized. As the suspended solid is consumed, the solution becomes clear.
Related Questions
Nitrogen monoxide is produced by combustion in an automobile engine. For the following reaction, 4.36 grams of nitrogen monoxide are mixed with excess oxygen gas . The reaction yields 5.46 grams of nitrogen dioxide . nitrogen monoxide ( g ) oxygen ( g ) nitrogen dioxide ( g ) What is the theoretical yield of nitrogen dioxide
Answers
Answer:
[tex]6.68~g~NO_2[/tex]
Explanation:
We have to start with the combustion reaction:
[tex]NO~+~O_2~->~NO_2[/tex]
Then we can balance the reaction:
[tex]2NO~+~O_2~->~2NO_2[/tex]
If we want to find the theoretical yield, we have to calculate the amount of [tex]NO_2[/tex]. To do this, we have to first convert the 4.36 g of [tex]NO[/tex] to moles [tex]NO[/tex] (using the molar mass 30 g/mol), then we have to convert from moles of [tex]NO[/tex] to moles of [tex]NO_2[/tex] (using the molar ratio) finally, we have to convert from moles of [tex]NO_2[/tex] to grams of [tex]NO_2[/tex] (using the molas mass 46 g/mol), so:
[tex]4.36~g~NO\frac{1~mol~NO}{4.36~g~NO}\frac{2~mol~NO_2}{2~mol~NO}\frac{46~g~NO_2}{1~mol~NO_2}=6.68~g~NO_2[/tex]
I hope it helps!
1.the particles in a_are packed very closely together. 2.the_in a liquid are loosely together. they can flow. 3.in a gas the_between the particles are bigger. if u re genius then answer this question class 4
Answers
Answer:
1. Solid
2. Particles.
3. Space.
Explanation:
1. The particles in a solid are packed very closely together. This is the most reason why solid particles have definite shapes of volume. This closeness is as a result of strong intermolecular forces force that exist between the particles. Hence, they can not flow but only vibrates about their mean position. Solids can not be compressed because of the strong intermolecular forces between their particles.
2. The particles of liquid are loosely together. In liquid, the particles are loosely packed and free to move about to certain degree. This is so because the intermolecular force between the particles are not as strong as those within the solid particles. Hence liquid has no definite shape but they have definite volume. They only assume the shape of the container they are poured into. Liquid can no be compressed..
3. In a gas, the space between the particles are bigger. This is so because the intermolecular forces between the particles are negligible i.e very small and so, the gas particles are free to move about and only restricted by the wall of the container they poured into. This negligible intermolecular forces are the reason why gas has no definite shapes and no volume. They can be compressed to fill a particular container.
A 2.950×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.7 mL . The density of water at 20.0∘C is 0.9982 g/mL.Calculate:
a. molality
b. mole fraction of glycerol in this solution
c. the concentration of the glycerol solution in percent by mass
d. the concentration of the glycerol solution in parts per million
Answers
Answer:
a. 2.959x10⁻²m
b. 5.327x10⁻⁴
c. 0.272%
d. 2718 PPM
Explanation:
A solution of 2.950x10⁻²M contains 2.950x10⁻² moles of Glycerol per Liter of solution. As the volume of the solution made was 1.000L, moles of glycerol are 2.950x10⁻².
a. molality: Molality is defined as the ratio between moles of solute (2.950x10⁻²) in kg of solvent. As there are 998.7mL of solvent and density is 0.9982g/mL, kg are:
998.7mL ₓ (0.9982g/mL) ₓ (1kg / 1000g) = 0.9969kg of solvent.
Molality: 2.950x10⁻² moles / 0.9969kg of solvent = 2.959x10⁻²m
b. Mole fraction is the ratio between moles of solute and total moles. Moles of water are:
998.7mL ₓ (0.9982g/mL) ₓ (1mol / 18.01g) = 55.35 moles of water.
Mole fraction glycerol:
2.950x10⁻² moles / (2.950x10⁻²moles + 55.35) = 5.327x10⁻⁴
c. Percent by mass Is the ratio by mass of solute and solution multiplied 100 times.
Mass of glycerol (Molar mass: 92.09g/mol):
2.950x10⁻² moles × (92.09g / mol) = 2.717g of glycerol
Mass of water:
998.7mL ₓ (0.9982g/mL) = 996.9g of water.
Percent by mass:
2.717g of glycerol / (996.9g of water + 2.717g) × 100 = 0.272%
d. Parts per million are defined as the ratio between mg of solute and kg of solution.
mg of 2.717g of glycerol are 2717mg
kg of solution are (996.9g + 2.717g) / 1000 = 0.9996kg
Parts per million:
2717mg / 0.9996kg = 2718 PPM
The reason for the dramatic decline in the number of measles cases from the 1960s to 2010 in the United States was because the vaccine
Answers
Answer:
It was because the vaccine generated actively acquired immunity, that is, inoculation of a portion of the measles virus so that the body forms the antibodies for a second contact and thus can destroy it without triggering the pathology.
Explanation:
Vaccines are methods of active acquired immunity since the antibody is not passively inoculated, it is manufactured by the body with a physiological process once part of the virus is inoculated.
The measles virus most of all affected the lives of infants or newborn children with severe rashes and high fevers that led to death.
In which of the following reactions will Kc = Kp? a. 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g) b. SO3(g) + NO(g) ⇌ SO2(g) + NO2(g) c. 2 N2(g) + O2(g) ⇌ 2 N2O(g) d. 2 SO2(g) + O2(g) ⇌ 2 SO3(g)
Answers
Answer:
The correct option is b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)
Explanation:
The relation between Kc and Kp is given by the following equation:
[tex]Kp = Kc (RT)^{dn}[/tex]
where R is the gas constant (0,082 L.atm/K.mol), T is the temperature (in K) and dn is the change in moles.
The change in moles (dn) is calculated as:
dn = moles of products - moles reactants
If dn=0, RT= 1 ⇒ Kc=Kp
We calculate dn for each reaction from the estequiometrial coefficients of products and reactants as follows:
a) 4 NH₃(g) + 3 O₂(g) ⇌ 2 N₂(g) + 6 H₂O(g)
dn= (2+6) - (4+3) = 1 ⇒ Kc ≠ Kp
b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)
dn = (1+1) - (1+1)= 0 ⇒ Kc = Kp
c) 2 N₂(g) + O₂(g) ⇌ 2 N₂O(g)
dn= 2 - (2+1) = -1 ⇒ Kc ≠ Kp
d) 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
dn = 2 - (2+1) = -1 ⇒ Kc ≠ Kp
The reaction in which Kc=Kp is b), because reactants and products have the same number of moles.
What is the percent composition of muscovite mica if its chemical formula is (KF)2 (Al2O3 )3 (SiO2 )6 (H2O)
Answers
Answer:
Explanation:
Hello,
To find the percentage composition of muscovite mica, we'll have to first find the molecular mass of the compound.
Chemical formula = (KF)₂(Al₂O₃)₃(SiO₂)₆(H₂O)
(KF)₂ = 58.097 × 2 = 116.194g/mol
(Al₂O₃)₃ = 3 × 101.96 = 305.88g/mol
(SiO₂)₆ = 6 × 60.08 = 360.48g/mol
H₂O = 18g/mol
(KF)₂(Al₂O₃)₃(SiO₂)₆(H₂O) = 116.194 + 305.88 + 360.48 + 18 = 800.554g/mol
Potassium = (78.18 / 800.554) × 100 = 9.765%
Fluorine = (38 / 800.554) × 100 = 4.75%
Aluminium = (162 / 800.554) × 100 = 20.23%
Silicon = (168.48/800.554) × 100 = 21.04%
Oxygen = (352/800.554) × 100 = 43.97%
Hydrogen = (2 / 800.554) × 100 = 0.24%
Muscovite mica is an aluminosilicate compound or a polysillicate compound found in rocks
What force is needed to accelerate a truck with mass 2,000 kg at a rate of 5.0
m/s22
O A. 1,000 N
B. 20,000 N
C. 2,000 N
O D, 10,000 14
Answers
Answer:
D
Explanation:
[tex]F=ma \\\\F=2000\cdot 5=10,000N[/tex]
Hope this helps!
Perform the conversions between energy units.
861 kJ=
J
3495 kcal=
kJ
7.84×106
Answers
Answer:
861 kJ = 861000 J,
3495 kcal = 14623.08 kJ
Explanation:
As I mentioned before, the last bit " 7.84 × 106 " just threw me off track, so I am simply going to assume that that does not appear in your question.
_______________________________________________________
Now we have 861 kilojoules, and have to convert it into joules for this first bit. Kilo being equal to 1000, to convert to joules you would have to multiply 861 by 1000, = 861000 Joules.
This second bit here asks us to convert 3495 kilocalorie to kilojoules. The difference between the two is that one is about 4.18 times greater than the other, so 3495 kilocalorie = 3495 * 4.18 = 14623.08 kilojoules.
Hope that helps!
Describe how you would prepare 150ml of 0.02N NaOH in the laboratory.
Answers
Answer:
Explanation:
molar mass of any subtance in 1 litre water gives 1N solution
for NaOH molar mass =40 g
40g --- 1000ml ---1N
Xg---- 150 ----0.02
X= 40*150*0.02/1000*1 =0.12 g
1. Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estimating their corresponding hydronium ion concentrations ([H3O+] using the ion product constant of water (Kw).
Kw = [H3O+][OH−] = [1×10−7 M][1×10−7 M] = 1×10−14 M
Hydronium ion concentration [H3O+] Solution condition
Greater than 1×10−7 M Acidic
Equal to 1×10−7 M Neutral
Less than 1×10−7 M Basic
Drag the appropriate items to their respective bins.
1. [OH−] = 6×10−12 M
2. [OH−] = 9×10−9 M
3. [OH−] = 8×10−10 M
4. [OH−] = 7×10−13 M
5. [OH−] = 2×10−2 M
6. [OH−] = 9×10−4 M
7. [OH−] = 5×10−5 M
8. [OH−] = 1×10−7 M
A. Acidic
B. Neutral
C. Basic
2. A solution has [H3O+] = 5.2×10−5M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [OH−] of the solution.
3. A solution has [OH−] = 2.7×10−2M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [H3O+] of the solution.
Answers
Answer:
Question 1.
1. [OH−] = 6×10−12 M is less than 1 * 10⁻⁷, therefore is acidic.
2. [OH−] = 9×10−9 M is less than 1 * 10⁻⁷, therefore is acidic.
3. [OH−] = 8×10−10 M is less than 1 * 10⁻⁷, therefore is acidic.
4. [OH−] = 7×10−13 M is less than 1 * 10⁻⁷, therefore is acidic.
5. [OH−] = 2×10−2 M is greater than 1 * 10⁻⁷, therefore is basic.
6. [OH−] = 9×10−4 M is greater than 1 * 10⁻⁷, therefore is basic.
7. [OH−] = 5×10−5 M is greater than 1 * 10⁻⁷, therefore is basic.
8. [OH−] = 1×10−7 M is equal to 1 * 10⁻⁷, therefore is neutral
Question 2:
[OH⁻] = 1.92 * 10⁻⁸ M
Question 3:
[H₃O⁺] = 3.70 * 10⁻¹¹ M
Explanation:
The ion product constant of water Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M² is a constant which gives the product of the concentrations of hydronium and hydroxide ions of dissociated pure water. The concentrations of the two ions are both equal to 1 * 10⁻⁷ in pure water.
A solution that has [OH⁻] greater than 1 * 10⁻⁷ is basic while one with [OH⁻] less than 1 * 10⁻⁷ is acidic.
1. [OH−] = 6×10−12 M is less than 1 * 10⁻⁷, therefore is acidic.
2. [OH−] = 9×10−9 M is less than 1 * 10⁻⁷, therefore is acidic.
3. [OH−] = 8×10−10 M is less than 1 * 10⁻⁷, therefore is acidic.
4. [OH−] = 7×10−13 M is less than 1 * 10⁻⁷, therefore is acidic.
5. [OH−] = 2×10−2 M is greater than 1 * 10⁻⁷, therefore is basic.
6. [OH−] = 9×10−4 M is greater than 1 * 10⁻⁷, therefore is basic.
7. [OH−] = 5×10−5 M is greater than 1 * 10⁻⁷, therefore is basic.
8. [OH−] = 1×10−7 M is equal to 1 * 10⁻⁷, therefore is neutral
Question 2:
Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[OH⁻] = 1 * 10⁻¹⁴ M²/ [H₃O⁺]
[OH⁻] = 1 * 10⁻¹⁴ M²/5.2*10⁻⁵ M
[OH⁻] = 1.92 * 10⁻⁸ M
Question 3:
Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴
[H₃O⁺] = 1 * 10⁻¹⁴ M²/ [OH⁻]
[H₃O⁺] = 1 * 10⁻¹⁴ M²/ 2.7 * 10⁻² M
[H₃O⁺] = 3.70 * 10⁻¹¹ M
The element silver has an atomic weight of 108 and consists of two stable isotopes silver-107 and silver-109. The isotope silver-107 has a mass of 107 amu and a percent natural abundance of 51.8 %. The isotope silver-109 has a percent natural abundance of 48.2 %. What is the mass of silver-109
Answers
Answer:
109
Explanation:
Let silver-107 be isotope A
Let silver-109 be isotope B
Let silver-107 abundance be A%
Let silver-109 abundance be B%
The following data were obtained from the question:
Atomic weight of silver = 108
Mass of isotope A (silver-107) = 107
Abundance of isotope A (silver-107) = 51.8%
Abundance of isotope B (silver-109) = 48.2%
Mass of isotope B (silver-109) =?
Now, we shall determine the mass silver-109 as follow:
Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]
108 = [(107 x 51.8)/100] + [(Mass of B x 48.2)/100]
108 = 55.426 + (Mass of B x 0.482)
Collect like terms
Mass of B x 0.482 = 108 – 55.426
Mass of B x 0.482 = 52.574
Divide both side by 0.482
Mass of B = 52.574/0.482
Mass of B = 109
Therefore, the mass of silver-109 is 109.
What is the daughter nucleus (nuclide) produced when Po210 undergoes alpha decay? Replace the question marks with the proper integers or symbols.
Answers
Answer:
Po - 206
Explanation:
Think about the nuclear equation for the alpha decay of Po - 210. Po - 210 has 84 protons and 126 neutrons in it's nucleus, and an alpha particle is identical to a helium - 4 nucleus, meaning it is present with 2 protons and 2 neutrons.
_______________________________________________________
Knowing that, during the alpha decay, Po - 210 will form a daughter nucleus identical to a polonium - 210 present with 4 less of it's atomic mass ( number of protons and neutrons ), and 2 less of it's atomic number ( number of protons ). Why? Because Po - 210 will split into this daughter nucleus and an alpha particle.
Po - 210 has an atomic mass of 210 and an atomic number of 84. It's daughter nucleus will have an atomic mass of 206, and an atomic number of 82, Po - 206.
Hope that helps!
The daughter nucleus (nuclide) produced when Po210 undergoes alpha decay is lead 206 decays by alpha emission to lead-206 (atomic number 82).
What is alpha decay ?The process of radioactive decay known as "alpha decay," in which an atomic nucleus produces an alpha particle and changes into a separate atomic nucleus with a mass number that is reduced by four and an atomic number that is reduced by two, is known as radioactive decay.
When a nucleus has too many protons, it becomes unstable and undergoes alpha decay. Energy and an alpha particle are released by the nucleus.
Po-210's atomic mass and atomic number both decrease by four after the emission of an alpha particle. The resulting atomic mass will be 2016, and the atomic number will be 82. This is an example of the lead isotope.
Thus, The daughter nucleus (nuclide) produced when Po210 undergoes alpha decay is lead 206.
To learn more about an alpha decay, follow the link;
https://brainly.com/question/2600896
#SPJ5
Dissolving NaOH (s) in water is exothermic. Two calorimetry experiments are set up.
Exp 1: 2 g of NaOH are dissolved in 100 mL of water
Exp 2: 4 g of NaoH are dissolved in 200 mL of water.
Which of the following is true?
A. both temperature changes will be the same
B. the second temeprature change will be approximately twice the first
C. the second temperature change will be approximately four times the first
D. the second temperature change will be approximately one-half of the first
E. the second temperature change will be approximately one-fourth the first
Answers
Answer:
Option-E
Explanation:
In the given question, the experiment was set up with 2 reactions in which reaction two has double the amount of reactants.
To calculate the heat released in experiments, the formula used will be
ΔH = mcΔT, where m= mass of
1. For reaction 1,
ΔH₁ = m₁c₁ΔT₁
2. For reaction 2,
ΔH₂ = m₂c₂ΔT₂
here, the concentration of each solute is the same but the amount is different. Reaction two has double the amount compared to the first reaction.
we compare ΔH₁ = 2 ΔH₂
m₁c₁ΔT₁= m₂c₂ΔT₂
here since concentration is the same, therefore cancelled out
m₁ΔT₁ = 2 m₂ΔT₂
m₁/2 x m₂ xΔT₁ = ΔT₂
100 / 2 x 200 ΔT₁
= 100/400 ΔT₁
= 1/4 ΔT₁= ΔT₂.
This shows that the second temperature change is 1/4 of the first.
Thus, Option-E is the correct answer.
Among three bases, X−, Y−, and Z−, the strongest one is Y−, and the weakest one is Z−. Rank their conjugate acids, HX, HY, and HZ, in order of decreasing strength. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them.
Answers
Answer: HZ > HX > HY in order of decreasing strengths.
Explanation: Generally, the rule is that the stronger the acid, the weaker its conjugate base and vice versa; same rule applies for bases and their conjugate acids.
So the weakest base Z- would have the strongest conjugate acid. Consequently, the strongest base Y- would have the weakest conjugate acid.
I hope this was MORE helpful as this is the correct answer.
The ranking of the conjugate acids in order of decreasing strength (i.e from strongest to weakest) is; HZ < HX < HY
First we must know that the stronger a base is, the weaker is it's conjugate acid and the weaker a base is, the stronger is it's conjugate acid.
Therefore, the order of decreasing strength of the conjugate acid is; HZ < HX < HY
Read more:
https://brainly.com/question/23917439
. A 0.100 M solution of the weak acid HA was titrated with 0.100 M NaOH. The pH measured when Vb = ½Ve was 4.62. Using activity coefficients, calculate pKa. The size of the A- anion is 450 pm.
Answers
Answer:
[tex]pk_a = 4.69[/tex]
Explanation:
[tex]V_b = \frac{1}{2} V_e[/tex]
[tex]pk_a(c) = 4.62\\pk_a(c) = - logk_a(c)\\4.62 = - logk_a(c)\\k_a(c) = antilog (-4.62)\\k_a(c) = 2.4 * 10^{-65}[/tex]
Activity coefficients:
[tex]f_A = 0.854\\f_H = 1.00\\f_{HA} = 1.00[/tex]
[tex]k_a = k_a(c) *f_A* \frac{f_H}{f_{HA}} \\k_a = 2.4 * 10^{-5} *0.854* \frac{1.0}{1.0}\\k_a = 2.05 * 10^{-5}[/tex]
[tex]pk_a = -log(k_a)\\pk_a = -log(2.05*10^{-5})\\pk_a = 4.69[/tex]
HELP PLEASE!
When phenol dissolves in water, a few (no all) of the molecules ionize H+ from the -OH group. This makes it
O a reducing agent
O an oxidizing agent
O a weak acid
a weak base
Answers
Answer: a weak acid
Explanation:
Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases. Oxidizing agent is the substance which itself gets reduced by gaining electrons and thus oxidize others.
Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced. Reducing agent is the substance which itself gets oxidized by losing electrons and thus reduces others.
Weak acids are those substances which dissociate partially to give [tex]H^+[/tex] ions when dissolved in water.
[tex]C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+[/tex]
Weak bases are those substances which dissociate partially to give [tex]OH^-[/tex] ions when dissolved in water.
[tex]NH_4OH\rightleftharpoons NH_4^++OH^-[/tex]
How are electrons arranged in molecules of compounds
Answers
Answer:
The electrons in an atom move around the nucleus in specific regions called electron shells. Every electron shell can just contain a specific number of electrons. The manner in which the electrons are arranged in an iota is known as the atom's electronic structure or electronic configuration of the specific atom.
Atoms or molecules are arranged in molecular compounds in set proportions by forming bonds, where outer shell valence electrons from each atom are shared between two same or different atoms of the compound.
Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2? 7. How many grams of ClF3 form from 130.0 grams of Cl2 when F2 is in excess? 8. How many grams of F2 are needed to react with 3.50 grams of Cl2?
Answers
Answer:
5) 1.147 moles Cl2
6) 12.57 grams ClF3
7) 339.10 grams ClF3
8) 5.63 grams F2
Explanation:
Step 1: Data given
Number of moles F2 = 3.44 moles
Molar mass F2 = 38.00 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles F2
For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3
For 3.44 moles F2 we'll need 3.44/3 = 1.147 moles Cl2
Step 1: Data given
Number of moles F2 = 0.204 moles
Molar mass F2 = 38.00 g/mol
Molar mass ClF3 = 92.448 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles ClF3
For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3
For 0.204 moles F2 we'll have 2/3 * 0.204 = 0.136 moles
Step 4: Calculate mass ClF3
Mass ClF3 = Moles ClF3 * molar mass ClF3
Mass ClF3 = 0.136 moles * 92.448 g/mol
Mass ClF3 = 12.57 grams ClF3
Step 1: Data given
Mass of Cl2 = 130.0 grams
Molar mass F2 = 38.00 g/mol
Molar mass ClF3 = 92.448 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles Cl2
Moles Cl2 = mass Cl2 / molar mass Cl2
Moles Cl2 = 130.0 grams / 70.9 g/mol
Moles Cl2 = 1.834 moles
Step 4: Calculate moles
For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3
For 1.834 moles Cl2 e'll have 2*1.834 = 3.668 moles ClF3
Step 5: Calculate mass ClF3
Mass ClF3 = Moles ClF3 * molar mass ClF3
Mass ClF3 = 3.668 moles * 92.448 g/mol
Mass ClF3 = 339.10 grams ClF3
Step 1: Data given
Mass of Cl2 = 3.50 grams
Molar mass F2 = 38.00 g/mol
Molar mass ClF3 = 92.448 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles Cl2
Moles Cl2 = Mass Cl2 / molar mass Cl2
Moles Cl2 = 3.50 grams / 70.9 g/mol
Moles Cl2 = 0.0494 moles
Step 4: Calculate moles F2
For 1 mol Cl2 we need 3 moles F2
For 0.0494 moles we need 3*0.0494 = 0.1482 moles
Step 5: Calculate mass F2
Mass F2 = moles F2 * molar mass F2
Mass F2 = 0.1482 moles * 38.00 g/mol
Mass F2 = 5.63 grams F2
A certain lightbulb containing argon at 1.20 atm and 18°C is
heated to 85°C at constant volume. Calculate its final pressure
(in atm).
Answers
Answer:
certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P 1 T 1 P 2 T 2 ... Ideal Gas Equation 5.4 Charles' law: V T (at constant n and P ) ... Consider a case in which two gases, A and B , are in a container of volume V.
Explanation:
Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 4.90 g of sulfuric acid and 4.90 g of lead(II) acetate are mixed. Calculate the number of grams of sulfuric acid, lead(II) acetate, lead(II) sulfate, and acetic acid present in the mixture after the reaction is complete .
Answers
Answer:
Mass H2SO4 = 3.42 grams
Mass of lead acetate = 0 grams
Mass PbSO4 = 4.58 grams
Mass of CH3COOH = 1.81 grams
Explanation:
Step 1: Data given
Mass of sulfuric acid = 4.90 grams
Molar mass of sulfuric acid = 98.08 g/mol
Mass of lead acetate = 4.90 grams
Molar mass of lead acetate = 325.29 g/mol
Step 2: The balanced equation
H2SO4 + Pb(C2H3O2)2 → PbSO4 + 2CH3COOH
Step 3: Calculate moles
Moles = mass / molar mass
Moles H2SO4 = 4.90 grams / 98.08 g/mol
Moles H2SO4 = 0.0500 moles
Moles lead acetate = 4.9 grams / 325.29 g/mol
Moles lead acetate = 0.0151 moles
Step 4: Calculate the limiting reactant
For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH
The limiting reactant is lead acetate. It will completzly be consumed (0.0151 moles). H2SO4 is in excess. There will react 0.0151 moles. There will remain 0.0500 - 0.0151 = 0.0349 moles
Step 5: Calculate moles of products
For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH
For 0.0151 moles lead acetate we'll have 0.0151 moles PbSO4 and 2*0.0151 = 0.0302 moles CH3COOH
Step 6: Calculate mass
Mass = moles * molar mass
Mass H2SO4 = 0.0349 moles * 98.08 g/mol
Mass H2SO4 = 3.42 grams
Mass PbSO4 = 0.0151 moles * 303.26 g/mol
Mass PbSO4 = 4.58 grams
Mass of CH3COOH = 0.0302 moles * 60.05 g/mol
Mass of CH3COOH = 1.81 grams
non-luminous flame is yellow.Explain
Answers
Answer:
It contains unburnt carbon particles of hot glowing solid carbon that gives out light
A friend asks you to help them decide which crackers are healthier. Comparing approximately equal serving sizes of 1 cracker (approx. 4.5 g Breton serving size vs. 4.7 g Triscuit), which would be a better choice with regards to calories, fat and sodium content?
a. Breton
b. Triscsuit
Answers
Answer:
The correct option is "Triscsuit"
Explanation:
In my opinion the correct option is tricsuit, because it has 0% saturated fat and TRANS fat, which is healthy fats since these fats are the worst for our body.
They also contain sodium but their levels are not high enough to trigger hypertension.
Substances like krypton, which is a gas at room temperature and pressure, can often be liquified or solidified only at very low temperatures. At a pressure of 1 atm, does not condense to a liquid until –153.2°C and does not freeze until –157.1°C. What are the equivalent absolute temperatures?
Answers
Answer:
The boiling and freezing temperatures of krypton at absolute scale are 119.95 K and 116.05 K, respectively.
Explanation:
The absolute temperature on SI units corresponds to Kelvin scale, whose conversion formula in terms of the Celsius scale is:
[tex]T_{K} = T_{C} + 273.15[/tex]
Where:
[tex]T_{K}[/tex] - Absolute temperature, measured in Kelvins.
[tex]T_{C}[/tex] - Relative temperature, measured in Celsius.
Finally, freezing and boiling temperatures are converted into absolute scale:
Boiling temperature
[tex]T_{K} = (-153.2 + 273.15)\,K[/tex]
[tex]T_{K} = 119.95\,K[/tex]
Freezing temperature
[tex]T_{K} = (-157.1 + 273.15)\,K[/tex]
[tex]T_{K} = 116.05\,K[/tex]
The boiling and freezing temperatures of krypton at absolute scale are 119.95 K and 116.05 K, respectively.
The lock-and-key model and the induced-fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Following are several statements concerning enzyme and substrate interaction. Indicate whether each statement is part of the lock-and-key model, the induced-fit model, or is common to both models.
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions
Answers
Answer:
The lock-and-key model:
c. Enzyme active site has a rigid structure complementary
The induced-fit model:
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
Common to both The lock-and-key model and The induced-fit model:
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
d. Substrate binds to the enzyme through non-covalent interactions
Explanation:
Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.
The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.
The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.
The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC2O4. The CaC2O4 weighed 40.3 g. What is the percent of calcium in the original sample
Answers
Answer:
128 gram of CaC2O4 contains 40 gram of Calcium
40.3 gram of CaC2O4 cotnains = 40*40.3/128 = 12.59 gram of Calcium
out of 15 gram 12.59 gram is Calcaim that means around 50% of orginal sample has Calcium
an auto of an element has 17 protons in its nucleus.a) write the electronic configuration of the atom.b)to what period and group does the element belong
Answers
Answer:
i hope it will help you
Explanation:
electronic configuration 1s²,2s,²2p^6,3s²3p^6,4s^1
as it has one electron in its valence shell so it is the member of group 1A(ALKALI METALS) and the number of shells is 4 so it is in period 4
Which 3 laws combine to make the Combined Gas Law?
Answers
Answer:
Boyle's Law, Charles' Law, and Gay-Lussac's Law
Explanation:
Answer:
Boyle's Law, Charles's Law, and Gay-Lussac's Law
Explanation:
Because cylinder with a volume of 3.00dm^3 contains 8.00 moles of oxygen gas at a temperature of 50.0 K. What is the pressure inside the cylinder? ( R=8.31)
Answers
Answer:
Option A. 1110 KPa.
Explanation:
The following data were obtained from the question:
Volume (V) = 3 dm³
Number of mole (n) = 8 moles
Temperature (T) = 50K
Gas constant (R) = 8.31 KPa.dm³/Kmol
Pressure (P) =..?
Pressure inside the cylinder can be obtained by using the ideal gas equation as follow:
PV =nRT
P x 3 = 8 x 8.31 x 50
Divide both side by 3
P= (8 x 8.31 x 50) /3
P = 1108 ≈ 1110 KPa
Therefore, the pressure inside the cylinder is 1110 KPa
When 18.0 mL of a 8.43×10-4 M cobalt(II) fluoride solution is combined with 22.0 mL of a 9.72×10-4 M sodium hydroxide solution does a precipitate form? no (yes or no) For these conditions the Reaction Quotient, Q, is equal to .
Answers
Answer:
Q = 1.08x10⁻¹⁰
Yes, precipitate is formed.
Explanation:
The reaction of CoF₂ with NaOH is:
CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).
The solubility product of the precipitate produced, Co(OH)₂, is:
Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)
And Ksp is:
Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²
Molar concentration of both ions is:
[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = 3.79x10⁻⁴M
[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = 5.35x10⁻⁴M
Reaction quotient under these concentrations is:
Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²
Q = 1.08x10⁻¹⁰
As Q > Ksp, the equilibrium will shift to the left producing Co(OH)₂(s) the precipitate
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g) 2.10 g were obtained. What is the percent yield of the reaction?
Answers
Answer:
[tex]Y=58.15\%[/tex]
Explanation:
Hello,
For the given chemical reaction:
[tex]Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)[/tex]
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
[tex]n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3[/tex]
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
[tex]n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3[/tex]
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
[tex]m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3[/tex]
Finally, we compute the percent yield with the obtained 2.10 g:
[tex]Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%[/tex]
Best regards.