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4 years ago

Which set of coefficients will balance the equation below? NaOH + Cu(NO3)2 Cu(OH)2 + NaNO3

Chemistry

2 answers:

Nonamiya [84]4 years ago

6 0

Answer is D
2 NaOH + 1Cu(NO3)2<span> equals 1Cu(OH)2 + 2NaNO3

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FinnZ [79.3K]4 years ago

5 0

The answer is D 2,1,1,2

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erica [24]

Answer:

1120 L.

Explanation:

Hello!

In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.

In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:

$V_2=\frac{V_1*n_2}{n_1}$

Thereby, we plug in to obtain:

$V_2=\frac{22.4L*50.0mol}{1mol}\\\\V_2=1120 L$

Best regards!

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3 years ago

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P

expeople1 [14]

Answer : The ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

Solution : Given,

$K_p=1.5\times 10^6$

$p_{O_2}$ = 0.21 atm

The given equilibrium reaction is,

$NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}$

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

$1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}$

$\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}$

$\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5$

Therefore, the ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

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3 years ago

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Mashcka [7]

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3 years ago

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natulia [17]

Answer:

Higher the frequency, the higher the energy

Explanation:

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3 years ago

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