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Nutka1998 [239]

3 years ago

6

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 496g of bromine liquid is produced

. Calculate the number of moles of bromine liquid produced.

1 answer:

vfiekz [6]3 years ago

3 0

Answer:

6.208 mol

Explanation:

Mass of Bromine Liquid = 496g

Number of moles = ?

Relationship between number of moles and mass is given as;

Number of moles = Mass / Molar mass

Molar mass of Bromine = 79.9g

Number of moles = 496 / 79.9 = 6.208 mol

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Arrange the following molecules according to the strength of their dispersion forces. Organize in order from largest to smallest

ki77a [65]

Explanation:

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

Therefore, more is the surface area occupied by the carbon chain more will be the dispersion forces present in it. Hence, less is the surface area occupied by a molecule less will be the dispersion forces present in it.

Hence, the given molecules are organized from largest to smallest dispersion forces as follows.

$CH_{3}CH(CH_{3})C(CH_{3})_{2}CH_{2}CH_{3}$ > $CH_{3}CH_{2}(CH_{2})_{4}CH_{2}CH_{3}$ > $CH_{3}CH_{2}CH_{2}CH_{2}CH_{3}$ > $CH_{3}C(CH_{3})2CH_{3}$ > $CH_{3}CH_{2}CH_{3}$ > $CH4$

3 0

3 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

$\frac{1}{22.4 L}\times 26 L= 1.1607 mol$

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

$\frac{Experimental}{Theoretical}\times 100$

$\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%$

60.42% is the percent yield of the reaction.

8 0

3 years ago

Hydrogen iodide is not produced by the same method as for hydrogen chloride.why??

lara [203]

Answer:

Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...

Using an ionic chloride and Phosphoric acid

H3PO4 + NaCl ==> HCl + NaH2PO4

H3PO4 + NaI ==> HI + NaH2PO4

H2SO4 + NaCl ==> HCl + NaHSO4

This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.

The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.

2I-² === I2 + 2e-

Explanation:

4 0

3 years ago

Hydrogenation reactions, in which H2 and an "unsaturated" organic compound combine, are used in the food, fuel, and polymer indu

ddd [48]

<u>Answer:</u> The amount of heat released is 56 MJ.

<u>Explanation:</u>

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

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

Given mass of $C_2H_6$ = 12 kg = 12000 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $C_2H_6$ = 30 g/mol

Putting values in above equation, we get:

$\text{Moles of }C_2H_6=\frac{12000g}{30g/mol}=400mol$

The chemical reaction for hydrogenation of ethene follows the equation:

$C_2H_4+H_2\rightarrow C_2H_6$

By Stoichiometry of the reaction:

When 1 mole of ethane releases 140 kJ of heat.

So, 400 moles of ethane will release = $\frac{140}{1}\times 400=56000kJ$ of heat.

Converting this into Mega joules, using the conversion factor:

1 MJ = 1000 kJ

So, $\Rightarrow 56000kJ\times (\frac{1MJ}{1000kJ})=56MJ$

Hence, the amount of heat released is 56 MJ.

8 0

3 years ago

What is the volume of a gas if 0.182 moles of the gas is at 1.99 atm and 83.4oC?

Pie

$\text{Given that,}\\\\\text{Pressure,}~ P = 1.99~ \text{atm}\\ \\\text{No. of moles,}~ n = 0.182~ \text{mol.}\\\\\text{Temperature,}~ T = 83.4^{\circ}~ C = 356.4~K\\\\ \text{Molar gas constant,}~ R = 0.082~ \text{L}~ \text{atm}~ \text{mol}^{-1}~ \text{K}^{-1}\\\\\text{Volume,}~ V =?\\\\\text{We know that,}~\\\\~~~~~~~PV =nRT\\\\\implies V = \dfrac{nRT}{P}\\\\\\~~~~~~~~~~=\dfrac{0.182 \times0.082 \times 356.4 }{1.99}\\\\\\~~~~~~~~~~\approx 2.68~ \text{L.}\\\\\\$

$\text{Hence the volume of the gas is 2.68 L.}$

6 0

2 years ago

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