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Setler [38]
4 years ago
14

How many grams of nitrogen are in a sample of Al(NO3)3 that contains 364.0 g of aluminum? Answer: ________ grams of nitrogen.

Chemistry
1 answer:
saw5 [17]4 years ago
7 0

<u>Answer:</u> The mass of nitrogen that is present for given amount of aluminium is 566.22 g.

<u>Explanation: </u>

We are given:

Mass of aluminium = 364 grams

The chemical formula of aluminium nitrate is Al(NO_3)_3

Molar mass of nitrogen = 14 g/mol

Molar mass of aluminium = 27 g/mol

In 1 mole of aluminium nitrate, 27 grams of aluminium is combining with 42 grams of nitrogen.

So, 364 grams of aluminium will combine with = \frac{42}{27}\times 364=566.22g of nitrogen.

Hence, the mass of nitrogen that is present for given amount of aluminium is 566.22 g.

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Which electron has a higher probability of being found at 4 Å of the nucleus, one in a 2s orbital or one in a 2p orbital?
Lena [83]

Answer:

one in a 2s orbital

Explanation:

Because of the peak near the nucleus in the 2s curve there is a higher probability of finding a 2s within 4 Å of the nucleus. In a multi-electron atom an electron in a 2s orbital will have a lower energy than one in a 2p orbital

4 0
3 years ago
In 1897 the Swedish explorer Andree tried to reach the North Pole in a balloon. The balloon was filled with hydrogen gas. The hy
vovangra [49]

Answer:

iron = 15000 kg 95% sulfuric acid = 27646.32 kg

Explanation:

The molar volume to a gas is 22.4 L, which means that a mole of a given gas at 0°C and 1 atm will have a volume os 22.4 L.

1 m3 - 1000 L

4870 m3 - x

x = 4870 * 1000 = 4.87 *10^6 L

1 mole - 22.4 L

y mole - 4.87 * 10^6 L

y = \frac{4.87 * 10^6}{22.4} = 2,17 *10^5 moles (if 100% of the H2 is used)

as 20% of it is lost, we need to know how much more we need to use.

80% - 2.17 * 10^5 moles

100% - z

z = \frac{2.17* 10^7}{80} = 2,68*10^5 moles

so we would need 2.68*10^5 moles of iron and sulfuric acid.

1 mole of iron - 56 *10^-3 kg

2.68*10^5 moles - t

t= 2.68*56 *10^2 = 15000 kg

1 mole of sulfuric acid - 98 * 10^-3 kg

2.68*10^5 - u

u = 2.68*98 *10^2 = 26264 kg (if 100%)

95% - 26264 kg

100% - v

v = \frac{26264 * 100}{95} = 27646.32 kg

6 0
3 years ago
How many moles of chlorine are used up in a reaction that produces 0.35kg of BCl3
maw [93]

4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.

(a) The <em>balanced chemical equation </em>is

2B + 3Cl2 → 2BCl3

(b) Convert kilograms of BCl3 to moles of BCl3

MM: B = 10.81; Cl = 35.45; BCl3 = 117.16

Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3

(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.

Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2

4 0
3 years ago
Guys help plzz assap
professor190 [17]
False should be your answer
8 0
3 years ago
Which statements differentiate DNA and RNA? Select three options.
Morgarella [4.7K]

Answer:

A.) DNA is a double helix, and RNA is a single strand.



C.) DNA is involved only in transcription, and RNA is involved in both transcription and translation.



E.) DNA does not have uracil as a nitrogen base, but RNA does have uracil as a nitrogen base.











Explanation:

Have a great rest of your day
#TheWizzer

3 0
2 years ago
Read 2 more answers
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