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3 years ago

Identify the variables in this hypothesis.

Chemistry

2 answers:

musickatia [10]3 years ago

7 0

Answer:

pressure and volume

Explanation:

Zanzabum3 years ago

5 0

Answer:

AJJJJJJJJJJJJJJ

Explanation:

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Based on the collision theory, which factors are likely to increase the rate of reaction? Select all that apply.

Alexus [3.1K]

Answer:

Option a

Explanation:

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2 years ago

You're legally obligated to yield the right-of-way:

kenny6666 [7]

It is most likely B.

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3 years ago

At which temperature would a can of soda be most fizzy when it is opened? 30°C 40°C 20°C 10°C

Helga [31]

The answer to your question is 40° because freezing temp for a liquid is 32°

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3 years ago

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2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

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3 years ago

What is the total pressure (in atm) inside of a vessel containing N2 exerting a partial pressure of 0.256 atm, He exerting a par

Bezzdna [24]

Answer: Total pressure inside of a vessel is 0.908 atm

Explanation:

According to Dalton's law, the total pressure is the sum of individual partial pressures. exerted by each gas alone.

$p_{total}=p_1+p_2+p_3$

$p_{N_2}$ = partial pressure of nitrogen = 0.256 atm

$p_{He}$ = partial pressure of helium = 203 mm Hg = 0.267 atm (760mmHg=1atm)

$p_{H_2}$ = partial pressure of hydrogen =39.0 kPa = 0.385 atm (1kPa=0.00987 atm)

Thus $p_{total}=p_{H_2}+p_{He}+p_{H_2}$

$p_{total}$ =0.256atm+0.267atm+0.385atm =0.908atm

Thus total pressure (in atm) inside of a vessel is 0.908

8 0

3 years ago

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