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Greeley [361]
3 years ago
10

Number 14? Please :)

Chemistry
1 answer:
LUCKY_DIMON [66]3 years ago
5 0

i believe the answer is because asphalt conducts heat better

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Ammonium perchlorate is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas , ch
FinnZ [79.3K]

Answer:

2.9 g of water are produced by 9.6 g of ammonium perchlorate

<em>Note: The question is incomplete. The complete question is given below:</em>

<em>Ammonium perchlorate (NH₄ClO₄) is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas (N₂) , chlorine gas (Cl₂), oxygen gas (O₂), water (H₂O) , and a great deal of energy. What mass of water is produced by the reaction of 9.6 g of ammonium perchlorate? Be sure your answer has the correct number of significant digits.</em>

Explanation:

The reaction of ammonium perchlorate (NH₄ClO₄) to produce nitrogen gas (N₂) , chlorine gas (Cl₂), oxygen gas (O₂), water (H₂O) is shown in the balanced chemical equation given below:

2 NH₄ClO₄ →  N₂ +  Cl₂ + 2 O₂ + 4 H₂O

From the equation, 2 moles of ammonium perchlorate produces 4 moles of water, i.e. mole ratio of NH₄ClO₄ to H₂O = 2 : 4 = 1 : 2

molar mass of ammonium perchlorate, NH₄ClO₄ = (14 + 4 * 1 + 35.5 +16 * 4) = 117.5

molar mass of water,  H₂O = (2 * 1 + 16) = 18.0 g

mass of water produced = moles of ammonium perchlorate * 2 * molar mass of water

moles of perchlorate = mass / molar mass = 9.6/117.5

mass of water produced = 9.6/117.5 * 2 * 18.0 g  = 2.94 g of water

Therefore, 2.9 g of water are produced by 9.6 g of ammonium perchlorate

6 0
3 years ago
For the reaction system, h2(g) + x2(g) ↔ 2 hx(g), kc = 24.4 at 300 k. a system made up from these components which is at equilib
Pavel [41]
Kc=24.4=[HX]∧2/[H2]×[X2] =(0.6)∧2/(0.2)×[H2]
[H2] = 0.36/(24.4×0.2) = 0.07377 mole
4 0
4 years ago
Read 2 more answers
40 cm3 of acid were mixed with 60 cm3 of alkali in an insulated container. What name is given to this type of reaction?
Stells [14]

Answer:

neutralization reaction

6 0
2 years ago
Which statements are correct relative to carbohydrates?
inna [77]

Answer:

Hi, the statements given in the question (a and b) are combination of right and wrong answers. But i will split the statement a and b into necessary fragments.

Carbohydrates are quick source of energy relatively. That is the only correct statement in the question.

Explanation:

Carbohydrates are naturally occurring organic compounds containing carbon, hydrogen and oxygen with the general formula (Cx H2y Oy) or Cx (H2O)y. They can be classified into simple sugar or complex sugar.

Simple sugars can be further divided into Monosaccharides (e.g glucose, fructose and galactose) and Disaccharides (e.g sucrose and maltose).

Complex sugar can be referred to as Polysaccharides (e.g cellulose and starch)

The statement above that says carbohydrate contains carbon,  hydrogen and nitrogen polymers is not correct.

Carbohydrates are not mostly monosaccharides. They also contain Disaccharides and Polysaccharides.

They are not the only source of fuel.

8 0
3 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
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