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3 years ago

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45.89 g of Hydrogen atoms contains how many moles of Hydrogen

1 answer:

Oksi-84 [34.3K]3 years ago

5 0

2.63 g

Explanation:

One way to approach this problem is to determine the percent composition of hydrogen in water.

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How many liters of carbon dioxide will 0.5 moles of lithium hydroxide (LiOH) absorb?

Elden [556K]

Answer : The volume of carbon dioxide will be, 5.6122 L

Solution : Given,

Moles of LiOH = 0.5 moles

Molar mass of carbon dioxide = 44 g/mole

Density of carbon dioxide = 0.00196 g/ml

First we have to calculate the mass of carbon dioxide.

The balanced reaction will be,

$2LiOH+CO_2\rightarrow Li_2CO_3+H_2O$

From the reaction we conclude that

As, 2 moles of LiOH absorbs 44 grams of carbon dioxide

So, 0.5 moles of LiOH absorbs $\frac{0.5moles}{2moles}\times 44g=11$ grams of carbon dioxide

Mass of carbon dioxide = 11 g

Density of carbon dioxide = 0.00196 g/ml

Now we have to calculate the volume of carbon dioxide.

$Density=\frac{Mass}{volume}$

$0.00196g/ml=\frac{11g}{volume}$

Volume of carbon dioxide = 5612.24 ml = 5.6122 L (1 L = 1000 ml)

Therefore, the volume of carbon dioxide will be, 5.6122 L

5 0

3 years ago

Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t

Vitek1552 [10]

<u>Answer:</u> The mass of decane produced is $1.743\times 10^2g$

<u>Explanation:</u>

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

$\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol$

The chemical equation for the hydrogenation of decene follows:

$C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)$

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = $\frac{1}{1}\times 1.225=1.225mol$ of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

$1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g$

Hence, the mass of decane produced is $1.743\times 10^2g$

5 0

3 years ago

A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mix

Alisiya [41]

Answer:

$M_f=38.8\%$

Explanation:

From the question we are told that:

Pressure $P=747mmHg$

Temperature $T=298K$

Volume $V=11.1$

Heat Produced $Q=780kJ$

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

$n= (747/760) *11.1/ (0.0821*298)$

$n=0.446mol$

Therefore

$x+y=0.446$

$x=0.446-y .....1$

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

$x(889) + y(2220) = 760 ...... 2$

Comparing Equation 1 and 2 and solving simultaneously

$x=0.446-y .....1$

$x(889) + y(2220) = 760 ...... 2$

$x=0.173$

$y=0.273$

Therefore

Mole fraction 0f Methane is mathematically given as

$M_f=\frac{x}{n}*100\%$

$M_f=\frac{1.173}{0.446}*100\%$

$M_f=38.8\%$

7 0

3 years ago

If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated

Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of water

i = Van't Hoff factor

$K_f$ = freezing point constant

m = molality

$w_b$ = mass of solute

$w_a$ = mass of solvent

$M_b$ = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0

3 years ago

Vodka is composed of 40% ethanol and 60% water. Water is the _____. while ethanol is the __, and vodka is a _____?

Mars2501 [29]

Answer:

The answer is A. solvent, solute, solution.

Explanation:

3 0

3 years ago