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2 months ago

If sodium is burned in chlorine gas, a compound is formed that dissolves in water. What colour will the solution be?

1 answer:

6 0

If sodium is burned in chlorine fuel, a compound is formed that dissolves in water. the solution be: Bright yellow mild

Chlorine is a yellow-green gas at room temperature. Chlorine has a smelly, annoying scent similar to bleach that is detectable at low concentrations. The density of chlorine gasoline is about 2.5 times extra than air, so one can reason it to initially stay near the floor in regions with little air movement.

Chlorine gasoline can be recognized by using its smelly, anxious smell, which is like the scent of bleach. The sturdy scent may additionally provide a good enough caution to human beings that they have been uncovered. Chlorine fuel appears to be yellow-green in color. Concentrations of approximately 400 ppm and past are commonly fatal over a half-hour, and at 1,000 ppm and above, fatality ensues within only some mins. A spectrum of scientific findings can be present in those uncovered to excessive tiers of chlorine.

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There are 34 questions on a test. John answers 22 correctly. What is Johns percent error?

VARVARA [1.3K]

Answer:

35 . 29%

Explanation:

no. of questions in test =34

no. of questions answered correctly =22

therefore, no. of questions answered incorrectly =34 - 22

=12

error percentage = 12/34 * 100

=35 .29 %

4 0

2 years ago

Acetylene gas, C2H2, can be produced by the reaction of calcium carbide and water. CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq

nekit [7.7K]

Answer:

1.0 L

Explanation:

Given that:-

Mass of $CaC_2$ = $2.54\ g$

Molar mass of $CaC_2$ = 64.099 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.54\ g}{64.099\ g/mol}$

$Moles_{CaC_2}= 0.0396\ mol$

According to the given reaction:-

$CaC_2_{(s)} + 2H_2O_{(l)}\rightarrow C_2H_2_{(g)} + Ca(OH)_2_{(aq)}$

1 mole of $CaC_2$ on reaction forms 1 mole of $C_2H_2$

0.0396 mole of $CaC_2$ on reaction forms 0.0396 mole of $C_2H_2$

Moles of $C_2H_2$ = 0.0396 moles

Considering ideal gas equation as:-

$PV=nRT$

where,

P = pressure of the gas = 742 mmHg

V = Volume of the gas = ?

T = Temperature of the gas = $26^oC=[26+273]K=299K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles = 0.0396 moles

Putting values in above equation, we get:

$742mmHg\times V=0.0396 mole\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\V=\frac{0.0396\times 62.3637\times 299}{742}\ L=1.0\ L$

<u>1.0 L of acetylene can be produced from 2.54 g $CaC_2$ .</u>

4 0

2 years ago

The state of matter in which a material has a definite shape and a definite volume is the _____ state.

Scilla [17]

Answer:

solid

Explanation:

in the solid state the material will has a fixed shape and volume whatever the container that contains it

where in liquid the shape will be different depending on the container

and in gas state the shape and volume are not definite

3 0

1 year ago

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Crazy boy [7]

1. The volume of ammonia consumed in the reaction is 23.2 L

2. The volume of oxygen consumed in the reaction is 29 L

<h3>Balanced equation</h3>

4NH₃(g) + 5O₂(g) --> 4NO(g) + 6H₂O(g)

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃ and 5 L of O₂

<h3>1. How to determine the volume of ammonia, NH₃ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 4) / 6 = 23.2 L of NH₃

Thus, 23.2 L of NH₃ were consumed

<h3>2. How to determine the volume of oxygen, O₂ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 5 L of O₂

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 5) / 6 = 29 L of O₂

Thus, 29 L of O₂ were consumed

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7 0

11 months ago

How many grams of NH3 can be produced from 2.35 mol of N2 and excess H2.

tatyana61 [14]

Since the molar ratio of N2 to Nh3 is 2:1, you can make twice as much NH3 by using N2. Therefore, you can make a maximum of 4.5 mol of NH3 assuming 100% yield.

6 0

2 years ago