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3 years ago

NaOH + H2SO4 =

1 answer:

3 0

$2NaOH+H_2SO_4\Rightarrow\ Na_2SO_4+2H_2O\\
NaOH+HNO_3\Rightarrow\ NaNO_3+H_2O\\
Cu(OH)_2+2HCl\Rightarrow\ CuCl_2+2H_2O\\
Cu(OH)2+H_2SO_4\Rightarrow\ CuSO_4+2H_2O\\
Cu(OH)_2+2HNO_3\Rightarrow\ Cu(NO_3)_2+2H_2O$

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3 years ago

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The outside temperature is 35C, what is the temperature in K? K=274.15+ c.

krek1111 [17]

The answer is 308 K.

The formula is C + 273.15 = K

this meaning the formula for this problem would be 35 + 273.15 = 308.15.

Out of all the option choices that would be rounded to 308K.

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3 years ago

At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti

mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

$rate=k[H_2]^2[NH_3]$

At a certain concentration of $H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]$

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

$0.120=k[H_2]^2[NH_3]$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[NH_3]$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}$

$\frac{R}{0.120}=\frac{1}{4}$

$R=0.03M/s$

Therefore, the initial rate of the reaction will be, 0.03 M/s

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3 years ago

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Answer:

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Explanation:

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