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Mkey [24]
3 years ago
13

How do you calculate the number of moles in CO2

Chemistry
1 answer:
gogolik [260]3 years ago
3 0

Answer:

The number of molecules in a mole (known as Avogadro's constant) is defined such that the mass of one mole of a substance, expressed in grams, is equal to the mean molecular mass of the substance. The molecular mass of CO2 = 12+2x16 = 44, so the mass of a mole of CO2 is approximalty 44 grams

Explanation:

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Answer: At the point when space experts take a gander at an article's range, they can decide its arrangement dependent on these frequencies. The most well-known technique stargazers use to decide the sythesis of stars, planets, and different articles is spectroscopy. This spread-out light is known as a range.

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How many atoms are there in each of the following ?a.)1.50 mol Na c.) 0.250 mol. Sib.)6.755 mol Pb
tamaranim1 [39]
a) \: n = \frac{N}{NA} \\ 1.5 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 1.5 \times 6.02 \times {10}^{23} \\ x = 9.03 \times {10}^{23} \: atoms \: of \: Na

b) \: n = \frac{N}{NA} \\ 0.25 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 0.25 \times 6.02 \times {10}^{23} \\ x = 1.505 \times {10}^{23} \: atoms \: of \: Si \:
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5 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
3 years ago
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