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2 years ago

Pls HELPPP

Chemistry

2 answers:

Yuri [45]2 years ago

8 0

maybe c

Explanation:

I'm not sure but maybe

kolezko [41]2 years ago

8 0

The molecules will always move from areas of higher concentration to areas of lower concentration when you are talking about diffusion.

The solution always want to be balanced, so if you have more, let’s say salt inside the cell, then the salt will flow through the membrane to the outer less salty solution until the salt level both inside and out the cell are equal.

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The combustion of 135 mg of a hydrocarbon produces 440 mg of CO2 and 135 mg H2O. The molar mass of the hydrocarbon is 270 g/mol.

NeX [460]

Answer:

Molecular formula = C20H30

Explanation:

NB 440mg = 0.44g, 135mg= 0.135g

From the question, moles of CO2= 0.44/44= 0.01mol

Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

Also from the question, moles of H2O = 0.135/18= 0.0075mole

Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H

To get the empirical formula, divide by smallest number of mole

Mol of C = 0.01/0.01=1

Mol of H = 0.015/0.01= 1.5

Multiply both by 2 to obtain a whole number

Mol of C =1×2 = 2

Mol of H= 1.5×2 = 3

Empirical formula= C2H3

[C2H3] not = 270

[ (2×12) + 3]n = 270

27n = 270

n=10

Molecular formula= [C2H3]10= C20H30

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3 years ago

If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr

victus00 [196]

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

$3NO_2+H_2O\rightarrow 2HNO_3+NO$

Moles of nitrogen dioxide :

$\frac{25.0 g}{56 g/mol}=0.5434 mol$

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

$\frac{2}{3}\times 0.5434 mol=0.3623 mol$ of nitric acid.

Mass of 0.3623 moles of nitric acid :

$0.3623 mol\times 63 g/mol=22.8260 g$

Theoretical yield = 22.8260 g

Experimental yield = ?

$\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100$

$75\%=\frac{\text{Experimental yield}}{22.8260 g}$

Experimental yield of nitric acid = 17.1195 g

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Answer:

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Answer:

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Answer:

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