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3 years ago

5.2 X 1026 molecules of CH4

Chemistry

1 answer:

olya-2409 [2.1K]3 years ago

8 0

<h3 /><h3>Answer above </h3>

I cant send this with no text 

Have a nice day

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Rubbing alcohol is a liquid that is clear and has a strong odor. these obstacles are called

dem82 [27]

The answer is D. Hope this helps.

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3 years ago

What is the relationship between absorbance and transmittance of a sample in a spectrophotometer?

Ulleksa [173]

I think Indirect, I’m so sorry if I’m wrong :(

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3 years ago

Given the following isotopes of strontium, calculate the average atomic mass.

vodka [1.7K]

Answer:

87.5198

Explanation:

(43.65 * 87.05) + (48.25 * 87.93) + (8.11 * 87.50) = 8751.98

8751.98 / 100 = 87.5198

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3 years ago

The rotten smell of a decaying animal carcass is partially due to a nitrogen-containing compound called putrescine. Elemental an

balu736 [363]

The empirical formula is $C_2H_6N$ .

Explanation:

Putrescine has the elements like Carbon, Nitrogen and Hydrogen present in them. So in order to determine the empirical formula, we first have to find the number of moles present in the putrescine. As the percentage of C, H and N present in the chemical is given as 54.50%, 13.73% and 31.77%, we assume that 100 g of Putrescine is taken as sample.

Then the mass of C, H and N present in Putrescine will be 54.50 g, 13.73 g and 31.77 g. We know that the molar mass of C is 12 g/mol, H is 1 g/mol and N is 14 g/mol. So divide the mass with the molar mass of the respective elements to determine the number of moles of these elements present in the sample.

$No.\ of\ moles\ of\ C=\frac{\text { Mass of } C}{\text { Molar mass of } C}=\frac{54.50 \mathrm{g}}{12 \mathrm{g} / \mathrm{mol}}=4.54\ moles$

Similarly, the number of moles of H and N present is determined.

$\text { No. of moles of } H=\frac{\text { Mass of } H}{\text { Molar mass of } H}=\frac{13.73 \mathrm{g}}{1 \mathrm{g} / \mathrm{mol}}=13.73 \text { moles }$

$No.\ of\ moles\ of\ N=\frac{\text { Mass of } N}{\text { Molar mass of } N}=\frac{31.77 \mathrm{g}}{14 \mathrm{g} / \mathrm{mol}}=2.27\ moles$

Then the empirical formula can be determined by dividing the number of moles of all elements with the least number of moles that is 2.27.

$\begin{aligned}&\text { No. of atoms of } C=\frac{4.54}{2.27}=2\\&\text { No. of atoms of } H=\frac{13.73}{2.27}=6\\&\text { No. of atoms of } N=\frac{2.27}{2.27}=1\end{aligned}$

So, the empirical formula is $C_2H_6N$ .

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3 years ago

Does anyone like having virtual school or naw?

Hoochie [10]

Answer:

Yesssss i really don't wanna go back to school

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3 years ago

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