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Rama09 [41]
3 years ago
12

Problem Page A chemist measures the amount of bromine liquid produced during an experiment. He finds that of bromine liquid is p

roduced. Calculate the number of moles of bromine liquid produced. Round your answer to significant digits.
Chemistry
1 answer:
masha68 [24]3 years ago
6 0

The question is incomplete, here is the complete question:

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.

<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of liquid bromine = 766. g

Molar mass of liquid bromine, (Br_2) = 159.8 g/mol

Putting values in above equation, we get:

\text{Moles of liquid bromine}=\frac{766.g}{159.8g/mol}=4.79mol

Hence, the amount of liquid bromine produced is 4.79 moles.

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A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O
ollegr [7]

Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) There is no other element present in the compound.

Explanation :

(a) Now we have to determine the masses of C and H in the sample.

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of CO_2=3.381g

Mass of H_2O=0.692g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, \frac{12}{44}\times 3.381=0.922g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, \frac{2}{18}\times 0.692=0.0769g of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) Now we have to determine the compound contain any other elements or not.

Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

This means that there is no other element present in the compound.

3 0
3 years ago
. The freezing point of an aqueous solution containing a nonelectrolyte solute is – 2.79 °C. What is the boiling point of this s
Semmy [17]

Answer:

Boiling point of the solution is 100.78°C

Explanation:

This is about colligative properties.

First of all, we need to calculate molality from the freezing point depression.

ΔT = Kf . m . i

As the solute is nonelectrolyte, i = 1

0°C - (-2.79°C) = 1.86 °C/m . m . 1

2.79°C / 1.86 m/°C = 1.5 m

Now, we go to the boiling point elevation

ΔT = Kb . m . i

Final T° - 100°C  =  0.52 °C/m . 1.5m . 1

Final T° =  0.52 °C/m . 1.5m . 1  + 100°C → 100.78°C

4 0
3 years ago
Determine the the mass of one molecule of hydrogen sulfide gas.
Alchen [17]

Answer:

the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.

6 0
3 years ago
A block of iron has a mass of 826 g. What is the volume of the block of iron whose density at 25°C is 7.9 ?
kolezko [41]

Answer:

104.56cm³

Explanation:

Mass=826g

Density =7.9g/cm³

Volume =Mass /Density

Volume =826/7.9

Volume =104.56cm³

5 0
2 years ago
do you think that religion and science are in conflict ,or that people can both have both without any problems
natka813 [3]
Religion and science are fundamentally incompatible. They disagree profoundly on how we obtain knowledge of the world. Science is based observation and reasoning from observation. Religion assumes that human beings can access a deeper level of information that is not available by either observation or reason.
5 0
3 years ago
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