![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:
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Answer:
Explanation:
the water make it sound different then above water
hop this helps
The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.
If 1 mole of vinegar contains 6.02 x 10^23 particles
x moles of vinegar contains 9.02 x 10^24 particles
x = 1 mole x 9.02 x 10^24 /6.02 x 10^23
x = 15 moles of vinegar
The reaction is as follows;
2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2
Since 2 moles of vinegar reacts with 1 mole of carbonate
x moles of vinegar reacts with 16.5 moles of carbonate
x = 2 moles x 16.5 moles/ 1 mole
x = 33 moles of vinegar
We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.
Theoretical yield = 16.5 moles x 158 g/mol = 2607 g
Actual yield = 6.35 moles x 158 g/mol = 1066.8 g
Percent yield = 1066.8 g/2607 g × 100/1
= 41%
Learn more: brainly.com/question/13440572?
Explanation:
Work done = force * perpendicular distance
= 1 * 0.01 = 0.01 joules
Answer : The correct answer for molarity of MgCl₂ solution is 0.67 M.
Molarity :
It expresses concentration of any solution . It can be defined as mole of solute present in volume of solution in Liter .
It uses unit M or 
It can be formulated as :
Given : mol of MgCl₂ (solute ) = 0.50 mol
Volume of solution = 750 mL
Following steps can be done :
Step 1 ) To convert volume of solution from mL to L
1 L = 1000 mL
Volume of solution = 0.750 L
Step 2) To find molarity
Plugging mole of solute and volume of solution in molarity formula as :
Hence molarity of MgCl₂ solution is 0.67 M
