![K_c = \frac{[NH_2]^{1}[H_2S]^{1}}{[NH_2HS]^{1}}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BNH_2%5D%5E%7B1%7D%5BH_2S%5D%5E%7B1%7D%7D%7B%5BNH_2HS%5D%5E%7B1%7D%7D)
is the equilibrium constant expression for
NH₂HS(S) → NH₂(g) + H₂S(g). Hence, option A is correct.
<h3>Definition of equilibrium constant.</h3>
A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.
Equilibrium constant for the NH₂HS(S) → NH₂(g) + H₂S(g) will be:
![K_c = \frac{[Product]^{coefficient}}{[Reactantt]^{coefficient}}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BProduct%5D%5E%7Bcoefficient%7D%7D%7B%5BReactantt%5D%5E%7Bcoefficient%7D%7D)
Hence, option A is correct.
Learn more about the equilibrium constant here:
brainly.com/question/10038290
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Answer:
Carbon-Carbon Bonds
Carbon can form single, double, or even triple bonds with other carbon atoms. In a single bond, two carbon atoms share one pair of electrons. In a double bond, they share two pairs of electrons, and in a triple bond they share three pairs of electrons
C. the square root of the mass of the particles.
<h3>Further explanation </h3>
Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or
the effusion rates of two gases = the square root of the inverse of their molar masses:
or
From this equation shows that the greater the mass of the gas, the smaller the effusion rate of the gas and vice versa, the smaller the mass of the gas, the greater the effusion velocity.
So if both gases are at the same temperature and pressure, the above formula can apply
Answer:
<h2>The first thing to do here is to use the molarity and the volume of the initial solution to figure out how many grams of copper(II) chloride it contains.</h2><h2 /><h2>133</h2><h2>mL solution</h2><h2>⋅</h2><h2>1</h2><h2>L</h2><h2>10</h2><h2>3</h2><h2>mL</h2><h2>⋅</h2><h2>7.90 moles CuCl</h2><h2>2</h2><h2>1</h2><h2>L solution</h2><h2>=</h2><h2>1.051 moles CuCl</h2><h2>2</h2><h2 /><h2>To convert this to grams, use the compound's molar mass</h2><h2 /><h2>1.051</h2><h2>moles CuCl</h2><h2>2</h2><h2>⋅</h2><h2>134.45 g</h2><h2>1</h2><h2>mole CuCl</h2><h2>2</h2><h2>=</h2><h2>141.31 g CuCl</h2><h2>2</h2><h2 /><h2>Now, you know that the diluted solution must contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride. As you know, when you dilute a solution, you increase the amount of solvent while keeping the amount of solute constant.</h2><h2 /><h2>This means that you must figure out what volume of the initial solution will contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride, the solute.</h2><h2 /><h2>4.49</h2><h2>g</h2><h2>⋅</h2><h2>133 mL solution</h2><h2>141.32</h2><h2>g</h2><h2>=</h2><h2>4.23 mL solution</h2><h2>−−−−−−−−−−−−−− </h2><h2 /><h2>The answer is rounded to three sig figs.</h2><h2 /><h2>You can thus say that when you dilute </h2><h2>4.23 mL</h2><h2> of </h2><h2>7.90 M</h2><h2> copper(II) chloride solution to a total volume of </h2><h2>51.5 mL</h2><h2> , you will have a solution that contains </h2><h2>4.49 g</h2><h2> of copper(II) chloride.</h2>
Explanation:
I think the answer is B ♂️ but I'm not sure
