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Y_Kistochka [10]
3 years ago
10

Can someone pls help me

Chemistry
2 answers:
Veronika [31]3 years ago
5 0

Answer:

22(don't forget to add the degrees

Explanation:

2x - 33 + 5x - 31 = 90

2x + 5x = 90 + 31 + 33

7x = 154

x = 154 \div 7

x = 22

Alchen [17]3 years ago
3 0

Answer:

A

Explanation:

That angle is a complementary angle which adds up to 90 degrees. So you write out the equation to solve for X. Equation= 2x-33+5x-31=90. Solve for x and you get 22 degrees.

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1.82 mol HCl ( 1 mol Cl2 / 4 mol HCl) = 0.46 mol Cl2

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Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi
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Answer : The value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

Explanation :

The following equilibrium reactions are :

(1) 2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2 K_1=5.40\times 10^{-16}

(2) 2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l) K_2=1.06\times 10^{10}

(3) CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g) K_3=2.68\times 10^{-9}

The final equilibrium reaction is :

CO_2(g)\rightleftharpoons C(s)+O_2(g) K_{goal}=?

Now we have to calculate the value of K_{goal} for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

K_{goal}=\sqrt{K_1\times K_2\times K_3}

Now put all the given values in this expression, we get :

K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}

K_{goal}=1.238\times 10^{-7}

Therefore, the value of K_{goal} for the final reaction is, 1.238\times 10^{-7}

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ohaa [14]
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