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2 years ago

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Calculate the internal energy of 2 moles of argon gas (assuming ideal behavior) at 298 K. Suggest two ways to increase its inter

nal energy by 10 J.

1 answer:

Dominik [7]2 years ago

5 0

I hope this helps you.

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A scientist measures the speed of sound in a monatomic gas to be 449 m/s at 20∘C. What is the molar mass of this gas?

Ivan

Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

To solve this problem, we can use the mathematical relation:

ν = $\sqrt{3RT/M}$

Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,

R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,

T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,

M is the molar mass of the gas in <em>(Kg/mol)</em>.

ν = $\sqrt{3RT/M}$

(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,

<em>by squaring the two sides:</em>

(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,

∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.

<em>∴ The molar mass of the gas is 36.25 g/mol.</em>

7 0

2 years ago

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

prisoha [69]

Answer:

2.76 × 10⁻¹¹

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹: 0 0 -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

$\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K$

T = (25.0 + 273.15) K = 298.15 K

$\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}$

3 0

2 years ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$

$7.00-6.86=log(\frac{base}{0.10})$

$0.14=log(\frac{base}{0.10})$

Taking antilog:

$10^0^.^1^4=\frac{base}{0.10}$

$1.38=\frac{base}{0.10}$

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is $Na_2HPO_4$ required to make the buffer is 0.138M.

5 0

2 years ago

A 220.0 gram piece of copper is dropped into 500.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what

FrozenT [24]

Answer:

C. 481 °C.

Explanation:

At equilibrium:

The amount of heat absorbed by water = the amount of heat released by copper.

To find the amount of heat, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of energy.

m is the mass of substance.

c is the specific heat capacity.

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T).

<em>∵ Q of copper = Q of water</em>

∴ - (m.c.ΔT) of copper = (m.c.ΔT) of water

m of copper = 220.0 g, c of copper = 0.39 J/g °C, ΔT of copper = final T - initial T = 42.00 °C - initial T.

m of water = 500.0 g, c of water = 4.18 J/g °C, ΔT of water = final T - initial T = 42.00 °C - 24.00 °C = 18.00 °C.

∴ - (220.0 g)( 0.39 J/g °C)(42.00 °C - Ti) = (500.0 g)(4.18 J/g °C)(18.00 °C)

∴ - (85.8)(42.00 °C - Ti) = 37620.

∴ (42.00 °C - Ti) = 37620/(- 85.8) = - 438.5.

∴ Ti = 42.00 °C + 438.5 = 480.5°C ≅ 481°C.

<em>So, the right choice is: C. 481 °C.</em>

4 0

2 years ago

Which ketone in each pair is more reactive?

KiRa [710]

Answer:

a. 2-heptanone is more reactive than 4-heptanone

b. chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone

Explanation:

The reactivity of the carbonyl compound (ketone ) is affected by the steric effect. The steric effect is a hindrance that occurs in the structure or reactivity of a molecule, which is affected by the physical size and the proximity of the adjacent parts of the molecule.

Between 2-heptanone or 4-heptanone, 2-heptanone is more reactive than 4-heptanone. This is because 2-heptanone is less affected by the steric hindrance, unlike the 4-heptanone.

Similarly, the reactivity of the carbonyl compound (ketone) is also affected by the polarity on the carbon compound, which is associated with how electronegative the substituent attached is to the carbonyl compound. From the periodic table, the electronegativity of the Halogen family decreases down the group. Therefore chlorine is more electronegative than bromine.

As such, chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone.

4 0

2 years ago