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2 years ago

5

Which best describes why acetic acid (CH3COOH) is an organic compound, but sodium hydroxide (NaOH) is not?

2 answers:

allsm [11]2 years ago

8 0

C) Acetic acid contains a significant amount of carbon, but sodium hydroxide contains no carbon

german2 years ago

3 0

The most appropriate answer is 'acetic acid contains a significant amount of carbon, but sodium hydroxide contains no carbon'. Organic compounds are those compounds that contains carbon atoms usually bonded to other elements such as oxygen, nitrogen and hydrogen. Organic compounds contain covalent single, double or even triple bonds.

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Will is learning about a new kind of plastic used to make models. He learns that infrared light is absorbed by the plastic, X-ra

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Answer:

no it dosen't matter

Explanation:

because it absorbed the light so it had to damage it.

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2 years ago

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If 5.61 g of ammonium nitrate (NH4NO3) is dissolved in 155 mL of water to make an instant cold pack, what is the molarity of the

kifflom [539]

Answer:

0.452 M

Explanation:

Remember that to get molarity, you have to go thru an equation:

M=mol/V (in liters)

So...

5.61 g NH4NO3 x 1 mol NH4NO3/80.043 g

=0.0701 mol NH4NO3

Convert mL to L

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Plug in and divide.

0.0701/0.155

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2 years ago

A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

mihalych1998 [28]

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= $\frac{4.00 g}{245 g/mol}=0.01633 mol$

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

$Molarity=\frac{n}{V(L)}$

Let the molarity of the solution be $M_1$

$M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M$

A students then takes 4 .00 mL of $M_1$ solution and dilute it to 275 ml.

$M_1=0.1633 M$

$V_1=4.00 mL$

$M_2=?$ (molarity after dilution)

$V_2=275 mL$ (after dilution)

$M_1V1=M-2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M$

Molarity of the of solution after dilution is 0.002375 M.

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

$[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M$

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions = $[NO_3^{-}]=0.004750 M$

$[NO_3^{-}]=\frac{n}{0.275 L}$

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097 grams of nitrate ions are there in the final solution.

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