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3 years ago

Cu + _ HNO₃ → ___ _Cu(NO₃)₂ + _ NO₂ + _ H₂O

Chemistry

2 answers:

Brilliant_brown [7]3 years ago

8 0

Answer:

the balanced equation is

2Cu + 8HNO₃ → 2Cu(NO₃)₂ + 4NO₂ + 4H₂O

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

ivolga24 [154]3 years ago

3 0

Answer:

Cu + 4HNO3 ---> Cu(NO3)2 + 2NO2 + 2H2O.

Explanation:

Balancing:

Cu + 4HNO3 ---> Cu(NO3)2 + 2 NO2 + 2H2O.

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

<u>Answer:</u> The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

<u>Initial:</u> 0.025 0.025

<u>At eqllm:</u> 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

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steposvetlana [31]

<h2>Answer:</h2>

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3 years ago

A student measures the mass of a sample as 9.64 g. Calculate the percentage error, given that the correct mass is 9.80 g.

rosijanka [135]

Answer:

<h3>The answer is 1.63 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual mass = 9.80 g

error = 9.80 - 9.64 = 0.16

We have

$p(\%) = \frac{0.16}{9.80} \times 100 \\ = 1.632653061...$

We have the final answer as

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4 years ago

__ Cu + ___ HNO₃ → ___ _Cu(NO₃)₂ + ___ NO₂ + ___ H₂O

Cu + _ HNO₃ → ___ _Cu(NO₃)₂ + _ NO₂ + _ H₂O