An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the molar mass (g/mol) of Q

Explanation:

A metal ion is a type of atom compound that has an electric charge.

Such atoms willingly lose electrons in order to build positive ions called cations. The selected  Ions are :

[tex]1. Mn^2^+\\\ 2. Ca^2^+\\\ 3. Co^2^+\\\ 4. Fe^2^-\\\ 5. K^+[/tex]

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

Answer:

As K >>> 1, the reaction will shift to the products

Explanation:

To know the direction of any reaction you must calculate the equilibrium constant, K. If K is < 1, the reaction will shift to the reactants and if k > 1 the reaction will shift to the products.

With the reactions:

HSO₄⁻ ⇄ SO₄²⁻ + H⁺ Ka = 1.26x10⁻²

And:

NH₄⁺ ⇄ NH₃⁺ + H⁺ Ka = 5.6x10⁻¹⁰

The inverse reaction:

NH₃⁺ + H⁺ ⇄ NH₄⁺ 1/Ka = 1.8x10⁹

The sum of the reactions:

HSO₄⁻ + NH₃⁺ + H⁺ ⇄ NH₄⁺ + SO₄²⁻ + H⁺ K = 1.26x10⁻² ₓ 1.8x10⁹ = 2.3x10⁷

Answer:

8.59 g

2.25 g

Explanation:

According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-

Moles of Pb(OH)CL is

[tex]= \frac{Mass}{Molar\ mass}[/tex]

[tex]= \frac{10.0 g}{259.65g / mol}[/tex]

= 0.0385 mol

Mass of PbO needed is

[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]

After solving the above equation we will get

= 8.59 g

Mass of NaCL needed is

[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]

After solving the above equation we will get

= 2.25 g

Therefore we have applied the above formula.

Answer:

[tex]\large \boxed{1.64\times 10^{-5}\text{ mol/L }}[/tex]

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

[tex]K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}[/tex]

The amount of the sample in space is referred to as concentration, in this the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

        Concentration of [tex]Na_2CO_3 = 0.00750 M[/tex]

           [tex](CO_3)^{2-} = Na_2CO_3 = 0.00750\ M\\\\Ksp \ \ Ag_2CO_3 =( Ag^{+})^2 (CO_3)^{2-}\\\\8.10 \times 10^{-12} = (Ag^+)^2 \times (0.00750\ M)\\\\(Ag^+)^2 = \frac{(8.10 \times 10^{-12})}{ (0.00750\ M)}\\\\(Ag^+)^2 = 1.08 \times 10^{-9}\ M^2\\\\(Ag^+) = (1.08 \times 10^{-9}\ M^2)^{\frac{1}{2}}\\\\\[(Ag^+)\] = (3.28 \times 10^{-5}\ M)\\\\[/tex]

So, the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].

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Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

The given question is incomplete, the complete question is:

Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol

A) –309 kJ/mol

B) –329 kJ/mol

C) None of the above

D) –349 kJ/mol

E) –369 kJ/mol

Answer:

The correct answer is option D, that is, -349 kJ/mol.

Explanation:

Based on the given information, the reaction is:  

NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.  

Therefore, there is a need to reverse the reaction and change the sign on ΔG.  

Now the reaction will become,  

Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).  

Answer:

intermolecular dipole-dipole hydrogen bonds

Explanation:

Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.

Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.

Answer:

Mn is manganese and CO₃ is carbonate. Since the charge for CO₃ is -2 and the subscript is 2, the charge of Mn must be +4 so the answer is manganese (IV) carbonate.

Manganese (IV) carbonate is the name of Mn(CO[tex]_3[/tex])[tex]_2[/tex]. The only names used to identify salts are those of the cation or the anion.

The chemical formula of the anion (such as chloride or acetate) comes first in the name of a salt, which is followed with the identity of the cation (such as sodium or ammonium). They are created when acids and bases react, and they are always composed of either metal cations or cations made of ammonium. Manganese is Mn, and carbonate is CO[tex]_3[/tex]. The solution equals manganese (IV) carbonate since the charge for CO[tex]_3[/tex] is -2 but the subscript is 2, meaning that the charge of Mn has to be +4.

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Answer:

[tex]Q=4154J[/tex]

Explanation:

Hello,

In this case, the involved heat in this heating process is considered to be computed via:

[tex]Q=nCp\Delta T[/tex]

Whereas we assume a constant molar specific heat of helium which is 20.77 J/(mol*K), thus, the transferred energy in the form of heat turns out:

[tex]Q=2mol*20.77\frac{J}{mol*K} *100K\\\\Q=4154J[/tex]

Regards.

Answer:

C.

Explanation:

Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.

Answer:

Option A. LR = N2O4, 45.7g N2 formed

Explanation:

The balanced equation for the reaction is given below:

N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and mass of N2 produced from the balanced equation. This is illustrated below:

Molar mass of N2O4 = 92.02 g/mol

Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02 g

Molar mass of N2H4 = 32.05 g/mol

Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g

Molar mass of N2 = 2x14.01 = 28.02g/mol

Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g

Summary:

From the balanced equation above,

92.02g of N2O4 reacted with 64.1g of N2H4 to produce 84.06g of N2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

92.02g of N2O4 reacted with 64.1g of N2H4.

Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.

From the calculations made above, we can see that only 34.83g out 45g of N2H4 is required to react completely with 50g of N2O4.

Therefore, N2O4 is the limiting reactant and N2H4 is the excess reactant.

Finally, we shall determine the mass of N2 produced from the reaction.

In this case the limiting reactant will be used as it will produce the maximum yield of N2 since all of it is used up in the reaction.

The limiting reactant is N2O4 and the mass N2 produced can be obtained as illustrated below:

From the balanced equation above,

92.02g of N2O4 reacted to produce 84.06g of N2.

Therefore 50g of N2O4 will react to produce = (50 x 84.06)/92.02 = 45.7g of N2.

Therefore, 45.7g of N2 were produced from the reaction.

At the end of the day,

The limiting reactant is N2O4 and 45.7g of N2 were produced from the reaction.

Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.

Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.

It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.

Answer:

B - Making an Observation

Explanation:

Making an observation best describes the act of using senses or tools to gather information. Therefore, option B is correct.

The five senses—sight, taste, touch, hearing, and smell—gather data about our surroundings that the brain interprets. Based on prior experience (and subsequent learning), as well as by combining the data from each sensor, we make sense of this information.

Information gleaned from your five senses is referred to as an observation. These are smell, taste, touch, hearing, and sight. When you see a bird or hear it sing, you notice it.

The term observation, which is also used to sense five aspects of the world including vision, taste, touch, smell, and hearing, is used to describe utilizing the senses to examine the world, employing tools to take measurements, and looking at prior research findings.

Thus, option B is correct.

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Answer:

THE NEW FREEZING POINT IS -4.196 °C

Explanation:

ΔTf = 1 Kf m

molarity of MgCl2:

Molar mass = (24 + 35.5 *2) g/mol

molar mass = 95 g/mol

7.15 g of MgCl2 in 100 g of water

7.15 g = 100 g

(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3

= 0.715 g /dm3

Molarity in mol/dm3 = molarity in g/dm3 / molar mass

= 0.715 g /dm3 / 95 g/mol

m = 0.00752 mol/ dm3

So therefore:

ΔTf = i Kf m

1 = 3 (1 Mg and 2 Cl)

Kf = 1.86 °C/m

M = 0.752 moles

So we have:

ΔTf = 3 * 1.86 * 0.752

ΔTf = 4.196 °C

The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C

Answer:

Explanation:

The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

            HCOOH ⇄ H⁺ + HCOO⁻

I               1.0M          0          0

C              -x            +x        +x

E            1.0-x            x          x

For the steps below, refer to the ICE chart above.

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

Answer:

a. both temperature changes will be the same

Explanation:

When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:

Q = m×C×ΔT

Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.

Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.

m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.

And Q is the heat released: If 2g release X heat, 4g release 2X.

Thus, ΔT in the experiments is:

Experiment 1:

X / 102C = ΔT

Experiment 2:

2X / 204C = ΔT

X / 102C = ΔT

That means,

Answer:

Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.

Explanation:

Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:

[tex]_{92} ^{238} U_{}[/tex]→ [tex]_{90} ^{234} Th_{} + _{2} ^{4} He_{}[/tex]

I hope this explanation is clear and explanatory.

Answer:

2.1 atm

Explanation:

Before we get the total pressure, we have to ensure all the gases have the same pressure unit.

Nitrogen gas = 0.75 atm

Helium = 490mmHg

To convert mmHg to atm;

760 mmHg = 1 atm

490 = x

x = 460 / 760 = 0.645 atm

Neon = 520 torr

Converting torr to atm;

760 torr = 1 atm

520 torr = x

x = 520 / 760 = 0.684 atm

The total pressure is then given as;

0.75 + 0.684 + 0.645 = 2.1 atm

Answer:

A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.

Explanation:

Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.

Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other  releases it.

As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.

Answer:

i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances

Explanation:

Answer:

None

Explanation:

Answer on Edge 2022

Answer:

(a) The empirical formula of the compound is

m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).

(b) The grams of O2 that were used in the reaction is 1.146 g

(c) The amount of O2 that would have been required for complete combustion is 1.401 g.

Explanation:

a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)

(b) Using law of conservation of mass from above

m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)

m(O2) = 0.446 + 0.700 + 0.430 - 0.430

m(O2) = 1.146 g

The grams of O2 that were used in the reaction is 1.146 g

(c) for complete combustion, we need to oxidized CO to CO2

Then, 2CO +O2 = 2CO2

m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}

m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g

Note; Molar mass of O2 = 32, CO = 28

m(total)(O2) = m(O2) + m(add)(O2)

m(total)(O2) = 1.146 + 0.255 = 1.401 g

The amount of that grams would have been required for complete combustion is 1.401 g.

Note (add) and (total) were used subscript to "m"

Answer:

Option A - When |ΔHsolute| > |ΔHhydration|

Explanation:

A solution is defined as a homogeneous mixture of 2 or more substances that can either be in the gas phase, liquid phase, solid phase.

The enthalpy of solution can either be positive (endothermic) or negative (exothermic).

Now, we know that enthalpy is amount of heat released or absorbed during the dissolving process at constant pressure.

Now, the first step in thus process involves breaking up of the solute. This involves breaking up all the intermolecular forces holding the solute together. This means that the solute molecules are separate from each other and the process is always endothermic because it requires energy to break interaction. Thus;

The enthalpy ΔH1 > 0.

Thus, the enthalpy of the solute has to be greater than the enthalpy of hydration.

An endothermic ΔHsolution occurs when |ΔH solute| < |ΔH hydration|.

A substance dissolves in water when the solute - solvent interaction exceeds the solute - solute solute interaction. The energy required to break the bonds between solutes is the ΔHsolute and the energy released when solute - solvent interaction take place is called the ΔHhydration.

We know that when  |ΔH solute| < |ΔH hydration|, energy is required to break up the solute - solute interaction and  ΔHsolution is endothermic.

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Answer:

8.33mL or .0083L

Explanation:

Use m1 * V1 = m2 * V2

6.00M(x) = 0.100M(500mL)

solve for x

x= (.1 * 500) / 6

x=8.333 mL

Answer:

Fronts

Explanation:

For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!

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Answer:

The answer is " 10.39"

Explanation:

Calculating acid moles:

[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]

Calculating NaOH moles:

[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]

calculating excess in OH-  Moles:

[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]

calculating total volume:

[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]

[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]

           [tex]=0.00025 M[/tex]

[tex]pOH = - \log 0.00025[/tex]

        = 3.6

[tex]pH = 14 - pOH[/tex]

      = 10.39

Answer:

[tex]\Delta S=6045.8\frac{J}{K}[/tex]

Explanation:

Hello,

In this case, we can compute the change in the entropy for vaporization processes in term of the enthalpy of vaporization as shown below:

[tex]\Delta S=\frac{m*\Delta H}{T}[/tex]

Whereas the temperature is in Kelvins. In such a way, the entropy results:

[tex]\Delta S=\frac{1.00kg*2256x10^3\frac{J}{kg} }{(100+273.15)K}\\\\\Delta S=6045.8\frac{J}{K}[/tex]

Best regards.

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}[/tex]

Answer:

Based on the difference in solubility one can perform the process of purification of the benzoic acid contaminated with sodium chloride. The benzoic acid does not get soluble in cold water, while the sodium chloride is soluble in cold water.  

Thus, for separation, the supplementation of cold water can be done into the mixture in the experiment of purifying benzoic acid from sodium chloride. In the process, the mixture is placed on the ice bath and is stirred well, in the end, the solution is filtered. The filtrate contains sodium chloride and on the filter paper pure benzoic acid is collected.