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3 years ago

X + Y XY + heat What happens as the temperature is increased? [X] remains constant. [X] increases. [X] decreases.

Chemistry

2 answers:

TEA [102]3 years ago

6 0

The given reaction is exothermic,

X + Y ----> XY + Heat

Heat is a product. So, if the temperature is increased, the equilibrium is shifted towards the reactants. So, the <u>[X] increases</u>.

Vilka [71]3 years ago

4 0

Answer: [X] increases.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$X+Y\rightarrow XY+heat$

This is a type of Exothermic reaction because heat is released in the reaction.

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in temperature occurs. As, this is an exothermic reaction, reverse reaction will decrease the temperature. Hence, the equilibrium will shift in the backward direction and thus [X] increases.

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3. Determine the moles of sodium, Na, containing 7.9x1024 atoms.

-BARSIC- [3]

Answer:

12.7mol Na.

Explanation:

Hello there!

In this case, according to the concept of mole, which stands for the amount of substance, we can recall the concept of Avogadro's number whereby we understand that one mole of any substance contains 6.022x10²³ particles, for the given atoms of sodium, we can calculate the moles as shown below:

$7.9x10^{23}atoms*\frac{1mol}{6.022x10^{23}atoms} \\\\$

Thus, by performing the division we obtain:

$12.7molNa$

Regards!

8 0

3 years ago

Nitrogen monoxide and water react to form ammonia and oxygen, like this: 4no (g) +6h2o (g) →4nh3 (g) +5o2 (g) the reaction is en

Archy [21]

The concept used here is the Le Chatelier's principle. When a disturbance is introduced to the system, it favors the direction of reaction that minimizes the disturbance to regain equilibrium.

In endothermic reactions, the forward reaction is favored when the temperature is low. Otherwise, the reverse reaction is favored. When you add the amounts of substances on the reactant side, more products would formed favoring the forward reaction. If you increase concentration on the product side, you form more reactants so it would favor the reverse reaction. Lastly, since 10 moles of gases are needed in the reactant side, it would be favored during high pressure reaction.

3 0

3 years ago

Numbers in scientific notation must be expressed in this form m times 10^n where

enot [183]

Answer:

Hi there I think the awnser is B. or A. lol but im 90% its A.

Explanation:

xoxo hope this helps

hugs and kisses also your pfp is hella cute:)

7 0

2 years ago

A friend tells you that plants grow and gain mass only because of the nutrients they pull out of the soil. It this statement tru

Nutka1998 [239]

The statement that the friend made is not true. most of the mass of the plant is from carbon. the carbon comes from carbon dioxide which is used during photosynthesis. the left over carbon from photosynthesis is used to to help the plant gain mass. there is a process for this which is called cellular respiration

5 0

3 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

8 0

3 years ago