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kogti [31]
3 years ago
12

Can u answer this? Mass: 126 g Volume: 15 cm to the third power Density:?

Chemistry
1 answer:
Andre45 [30]3 years ago
4 0

Answer:

Calculate density

Weight/mass

126

g

Volume

15

cm³

Density

8,400

kg/m³

Explanation:

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four atoms of oxygen

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3 years ago
7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s
Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both NH_{3} and CO_{2} are gaseous. Hence equilibrium constant depends upon partial pressures of NH_{3} and CO_{2}.

Initially no NH_{3} and CO_{2} were present.

Hence mole fraction of NH_{3} and CO_{2} at equilibrium can be calculated from coefficient of NH_{3} and CO_{2} in balanced equation.

Mole fraction of NH_{3} = (number of moles of NH_{3})/(total number of moles of NH_{3} and CO_{2}) = \frac{2moles}{(2+1)moles}=\frac{2}{3}

Mole fraction of CO_{2} = (number of moles of CO_{2})/(total number of moles of NH_{3} and CO_{2}) = \frac{1moles}{(2+1)moles}=\frac{1}{3}

Let's assume both CO_{2} and NH_{3} behaves ideally.

Therefore partial pressure of NH_{3}, P_{NH_{3}}= x_{NH_{3}}.P_{total} and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm

P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm

So, K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888

4 0
3 years ago
16. The concentration of a solution of potassium hydroxide is determined by titration with nitric
____ [38]

Answer:

M_{base}=0.709M

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

KOH+HNO_3\rightarrow KNO_3+H_2O

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

n_{acid}=n_{base}

That in terms of molarities and volumes is:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, solving the molarity of the base (KOH), we obtain:

M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M

Regards.

3 0
3 years ago
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