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Usimov [2.4K]
3 years ago
7

If you were asked to convert 25 mg to the unit hg, which of the following would be the first fraction used in the conversion? te

n to the negative third power hg over one g ten to the negative third power mg over one hg ten to the negative third power mg over one g ten to the negative third power g over one mg
Chemistry
1 answer:
NemiM [27]3 years ago
6 0
The conversion factor is added to the original given unit so that you can end up with the final unit. Basically, the conversion unit does not change the value because the factor is just equal to 1. You just manipulate the units by cancelling them.

Among the given choices, the fraction which is equal to 1 is 10⁻³ g/ 1 mg, because there are 1,000 mg per 1 g. In reverse, that would be 10⁻³ of a gram.
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The reaction: 2 CO(l) + O2(g) ⇄2 CO2(g) with a H = −25.0 kJ/mol, is at equilibrium. Which of the following lists three ways thi
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Option B and D both have 3 ways to shift the reaction to the right

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The reaction  2 CO(l) + O2(g) ⇄2 CO2(g) is exothermic because ΔH = -25 kJ < 0 This means heat will be released

Therefore, increasing the temperature will shift the equilibrium to the left, while decreasing the temperature will shift the equilibrium to the right.

A. remove CO2, lower pressure and remove CO

⇒ Lowering the pressure will shift the equilibrium to the side with most moles of gas : On the left side there are 3 moles, on the right side 2 moles. By lowering the pressure, the equilibrium will shift to the<u> left.</u>

B  remove CO2, raise pressure and add CO

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the <u>right</u> side.

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The<u> right</u> side).

⇒ Add CO: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added CO. (The<u> right</u> side).

C raise temperature, lower pressure and remove O2

⇒ Increasing the temperature will shift the equilibrium to the <u>left</u>

D add O2, raise pressure and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the <u>right</u>

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the <u>right</u> side.

⇒ Add O2: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added O2. (The<u> right</u> side).

E  remove CO2, increase volume and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the <u>right</u>

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The <u>right</u> side).

⇒ Increase volume : with a pressure decrease due to an increase in volume, the side with more moles is more favorable. The equilibrium will shift to the <u>left</u> side.

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