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JulijaS [17]
3 years ago
10

How many moles of N are in 0.241 g of N2O?

Chemistry
2 answers:
gtnhenbr [62]3 years ago
6 0
0.00547 moles of N.
eduard3 years ago
6 0

Answer:

Explanation:

Calculate the molar mass of the N_{2}O:

Molar mass N = 14\frac{g}{mol}

Molar mass O = 16\frac{g}{mol}

Molar mass N_{2}O = 2*(Molar mass of N) + Molar mass of O

Molar mass N_{2}O = 2*(14\frac{g}{mol})+16\frac{g}{mol}

Molar mass N_{2}O = 44\frac{g}{mol}

Calculate the number of moles of N in 0.241g of N_{2}O:

0.241gN_{2}O*\frac{1molN_{2}O}{44gN_{2}O}*\frac{2molesN}{1molN_{2}O}=0.01molesN

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The Nutrition Facts label for crackers states that one serving contains 19 g of carbohydrate, 4 g of fat, and 2 g of protein. Wh
sp2606 [1]

<u>Answer:</u> The amount of kilocalories contained in the given serving of crackes is 0.120 kCal

<u>Explanation:</u>

<u>As per the USDA:</u>

Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram

We are given:

Mass of fat in the crackers = 4 g

Mass of carbohydrates in the crackers = 19 g

Mass of protein in the crackers = 2 g

Conversion factor used:  1 kCal = 1000 Cal

Applying unitary method:

  • <u>For Fat:</u>

1 gram of fat provides 9 calories

So, 4 gram of fat will provide = \frac{9}{1}\times 4=36Cal=0.036kCal

  • <u>For Carbohydrates:</u>

1 gram of carbohydrates provides 4 calories

So, 19 gram of carbohydrates will provide = \frac{4}{1}\times 19=76Cal=0.076kCal

  • <u>For Proteins:</u>

1 gram of proteins provides 4 calories

So, 2 gram of fat will provide = \frac{4}{1}\times 2=8Cal=0.008kCal

Total kilocalories per serving = [0.036 + 0.076 + 0.008] kCal = 0.120 kCal

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Which tool is used to measure mass?
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8. Consider the element Scandium, atomic # 21.
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Explanation:

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The final electron is placed in the d orbital. The shell is 3d

(b) When scandium has a charge if +1, it has lost an electron. The total number of electrons would now be 21-1 = 20

The electronic configuration would be given as;

1s2 2s2 2p6 3s2 3p6 4s2

The electron in the 3d orbital would be removed.

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