What would happen to a system at equilibrium if the temperature were

Chemistry

2 answers:

igomit [66]3 years ago

7 0

Answer: D. The equilibrium constant would change.

Irina18 [472]3 years ago

5 0

Answer:

It’s d

Explanation:

Yep

You might be interested in

A neutral particle formed when atoms share electrons.

Yakvenalex [24]

Answer:

ion

ionic bond

compound

convalent bond

electrons

molecule

Explanation:

6 0

3 years ago

Mercury can be obtained by reacting mercury(I) sulfide with calcium oxide. How many grams of calcium oxide are needed to produce

OLga [1]

First, 55 g of Hg is 3.65 moles because one mole of Hg has a molar mass of 200.59

Then, the mole ratio of Hg to CaO is 8:4 or 2:1. SO we divide 3.65 by 2 to get 1.82 moles of CaO

This is the same as 102.06 grams because one mole of CaO has a molar mass of 56.0774

Hope this helps!

4 0

3 years ago

A hot air balloon has an air vent that keeps the air pressure inside and outside the same. Allen observes that a hot air balloon

DedPeter [7]

Answer:

B - The high temperature makes the gas molecules spread apart according to Charles's law because this law describes how a gas will behave at constant pressure.

Explanation:

Charle's Law describes the relationship between temperature and volume, where increased temperature leads to increased volume. When volume is increased, that means the gas molecules are more spread apart and have more random motion. Therefore, the answer is B.

6 0

3 years ago

"Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Alex17521 [72]

Answer:

$\boxed{\text{0.50 mol/L}}$

Explanation:

The balanced equation is

2COF₂ ⇌ CO₂+CF₄; Kc = 9.00

1. Set up an ICE table

$\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}$

2. Solve for x

$K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}$