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AfilCa [17]
3 years ago
8

_Zn+_HCLarrow_ZnCl2+_H2

Chemistry
1 answer:
dsp733 years ago
6 0
Plug in any number that yo think it’s the answer and make a t chart and put every letter in it and the other side the number until it is balanced
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(a) calculate the molarity of a solution made by dissolving 0.0815 mol na2so4 in enough water to form exactly 550. ml of solutio
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You just need to convert it into moles per Liter (mol/L).

0.0815mol / 0.550L = 0.148mol/L
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Why could a diet low in carbohydrates be dangerous?
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B-The body could break down proteins from muscles for energy.
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Zac releases the air from a balloon. Which statement best describes what will happen to the air that was inside the balloon?
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The correct answer is the last option. When <span>Zac releases the air from a balloon, it will expand to fill the room. It will not expand in the sense that molecules will be big but the molecules will be more spread out into the room and more air molecules will be present to fill the room.</span>
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At what temperature does uranium hexafluoride have a density of 0.9560 g/L at 0.5073 atm?
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Density = 0.7360 g/L.

Pressure = 0.5073 atm.
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Chemistry homework question answer, step 2, image 1
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Here, R is the universal gas constant (0.0821 L-atm/mol K), T is the temperature in Kelvin, and n is the number of
8 0
3 years ago
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:

<u>Chemical reaction involved</u>: 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

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

\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)

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

4 0
3 years ago
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