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Anit [1.1K]
3 years ago
10

When a 1.00-g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3°C. When

Chemistry
1 answer:
Advocard [28]3 years ago
3 0

Answer:

The energies of  combustion (per gram) for hydrogen and methane are as follows: Methane = 82.5 kJ/g;  Hydrogen = 162 kJ/g

<em>Note: The question is incomplete. The complete question is given below:</em>

To compare the energies of combustion of these fuels, the  following experiment was carried out using a bomb  calorimeter with a heat capacity of 11.3 kJ/℃.  When a 1.00-g sample of methane gas burned with

<em>excess oxygen in the calorimeter, the temperature  increased by 7.3℃. When a 1.00 g sample of  hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Compare the energies of  combustion (per gram) for hydrogen and methane.</em>

Explanation:

From the equation of the first law of thermodynamics, ΔU = Q + W

Since there is no expansion work in the bomb calorimeter,  ΔU = Q

But Q = CΔT

where C is heat capacity of the bomb calorimeter =  11.3
kJ/ºC; ΔT = temperature change

For combustion of methane gas:

Q per gram = (
11.3
kJ/ºC * 7.3°C)/1.0g

Q = 83 kJ/g

For combustion of hydrogen gas:

Q per gram = (
11.3
kJ/ºC * 14.3°C)/1.0g

Q = 162 kJ/g

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What is the empirical formula for the following molecular formula: C10H5O2
Tju [1.3M]

The empirical formula is the same as the molecular formula : C₁₀H₅O₂

<h3>Further explanation</h3>

Given

Molecular formula : C₁₀H₅O₂

Required

The empirical formula

Solution

The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.  

The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

<em>(EF)n=MF </em>

(EF)n = C₁₀H₅O₂

If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula

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3 years ago
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Vinil7 [7]

Answer:

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Explanation:

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Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what
Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

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3 years ago
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A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the vol
Troyanec [42]

Answer : The correct answer for molarity = 0.19 \frac{mol}{L} and Molality = 0.18 \frac{mol}{Kg}.

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}\

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}

Molar mass of aniline (C₆H₅NH₂) = 93.13 \frac{g}{1 mol}

Mass of aniline = 6.2 g

Plugging values in mole formula :

Mole  =  \frac{6.2 g}{93.13 \frac{g}{1 mol}}

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

Volume of solution = \frac{350 mL}{1000mL}  * 1 L

Volume of solution = 0.350 L

Step 3: Plug value of mole and volume in molarity formula :

Molarity = \frac{0.0665 mol}{0.350 L }

Molarity = 0.19 M or 0.19 \frac{mol}{L}

------------------------------------------------------------------------------------------

2) Molality : It is defined as mole of solute present in Kilogram of solvent . It can be expressed as :

Molality (m) = \frac{mole of solute (mol)}{Kilogram of solvent (Kg)}

Following are the steps to calculate molality :

Step 1: To find mole of Solute :

Mole of solute can be found out using mole formula . It is same as done for molarity .

Mole = 0.0665 mol

Step 2 : To find kilogram of solvent :

Mass of solvent can be calculated using density formula as :

Density \frac{g}{mL} = \frac{mass (g) }{volume (mL)}

Plugging value in density formula :

1.04 \frac{g}{mL}  =  \frac{ mass }{350 mL}

Multiplying both side by 350 mL

1.04 \frac{g}{mL} * 350 mL = \frac{x}{350 mL } * 350 mL

Mass of solvent = 364 g

Since mass is in g, it need to be converted to Kg . ( 1 Kg = 1000 g )

Mass of solvent = \frac{364 g}{1000g} * 1 Kg

Mass of solvent = 0.364 Kg

Step 3: Plug values of mole and Kg in molality formula :

Molality = \frac{0.0665 mol}{0.364 Kg}

Molality = 0.18 m or 0.18 \frac{mol}{Kg}

3 0
3 years ago
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