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3 years ago

From what are chemical sedimentary rocks formed?

Chemistry

2 answers:

mina [271]3 years ago

8 0

Chemicals cemented together

masha68 [24]3 years ago

6 0

Chemicals dissolved in water. Calcite is a good example, if I'm not mistaken.

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If 9.9L of helium are in a tire at 303k,how many liters will be present at 403k if the pressure is held constant.

Slav-nsk [51]

Answer:

Final volume, V2 = 13.18 Liters

Explanation:

Given the following data;

Initial volume = 9.9 L

Initial temperature = 303 K

Final temperature = 403 K

To find the final volume, we would use Charles law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

$VT = K$

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

Where;

V1 and V2 represents the initial and final volumes respectively.
T1 and T2 represents the initial and final temperatures respectively.

$\frac{V1}{T1} = \frac{V2}{T2}$

Making V2 as the subject formula, we have;

$V_{2}= \frac{V1}{T1} * T_{2}$

$V_{2}= \frac{9.9}{303} * 403$

$V_{2}= 0.0327 * 403$

Final volume, V2 = 13.18 Liters

8 0

3 years ago

A glass breaks is a physical or chemical change

maksim [4K]

Breaking glass is an example of a physical change because if you break it, it is still glass.

7 0

3 years ago

Read 2 more answers

26. A Grignard's reagent may be made by

Marat540 [252]

Its pretty easy, correct answer is D

8 0

3 years ago

Does more data mean more reliable results or less

Dvinal [7]

Answer:

more reliable. The more results the better results you get.

Explanation:

6 0

3 years ago

Read 2 more answers

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago