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3 years ago

What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?

1 answer:

katrin2010 [14]3 years ago

6 0

As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane

Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ

X = 0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol

Mass = 110.83 g of Octane

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How much heat is required to raise the temperature of 250.0g of mercury 52.0°c if the specific heat of mercury is 0.140 j/(g x °

sasho [114]

Q = m c T

c= 0.140 j/(g x °c)

m= 250.0g

T =52

hope you can solve it now

8 0

3 years ago

Acid rain can be destructive to both the natural environment and human-made structures. The equation below shows a

guapka [62]

Answer:

Explanation:

What the equation is tell you is that for every 3 mols of NO2 you get 2 mol of HNO3

3 mol NO2 / 2 mol HNO2 ===> 300.00 mol NO2 / x Cross multiply

3x = 2 * 300

3x = 600 Divide by 3

3x/3 = 600/3 Do the division

x = 200.00

3 0

3 years ago

Calculate the concentration of h3o⁺ in a solution that contains 5.5 × 10-5 m oh⁻ at 25°c. Identify the solution as acidic

Andreas93 [3]

Answer: $[H_3O^+]=1.6\times 10^{-10}M$ , the solution is basic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

$pH=-\log [H_3O^+]$

$pOH=-\log[OH^-]$

Putting in the values:

$pOH=-\log[5.5\times 10^{-5}]$

$pOH=4.2$

$pH+pOH=14$

$pH=(14-pOH)=(14-4.2)=9.8$

$9.8=-\log [H_3O^+]$

$[H_3O^+]=1.6\times 10^{-10}M$

As pH is more than 7, the solution is basic

7 0

3 years ago

Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb=4.75. Calculate the pH of 1.0 L of the so

vladimir2022 [97]

You have 0.50 mol of NH3 and 0.20 mol of NH4+ to start (NH4Cl dissolves completely), given the molarity and 1.0 L solution.

30.0 mL of 1.0 M HCl is 0.0300 mol of HCl. This will react with the NH3 to produced 0.030 mol of NH4+.

You now have 0.47 mol NH3 and 0.23 mol NH4+. Now use the Henderson-Hasselbach equation to calculate your pH. The equation says to use concentration of acid and base, but you can just use the moles of them because it doesn’t make a difference.

pH = pKa + log(base/acid)

pKa = 14 - pKb = 14 - 4.75 = 9.25

pH = 9.25 + log(0.47/0.23) = 9.56

5 0

3 years ago

Hydrogen sulfide gas is combusted with oxygen gas to produce sulfur dioxide gas and water vapour. If there is 48.4 L of oxygen a

Kitty [74]

Answer:

2H2S + 3O2 → 2SO2 + 2H2O

V(O2) = 48.4 L

p = 105 kPa = 1.036 atm

T = 190 + 273 = 463 K

Ideal gas law:

pV = nRT

n = \frac{pV}{RT}n=

R = 0.08206 L×atm/mol×K

n(O2) = \frac{1.036 \times 48.4}{0.08206 \times 463}=1.319 \; mol=

0.08206×463

1.036×48.4

=1.319mol

According to the reaction:

n(H2S) = \frac{2}{3}

n(O2) = \frac{2}{3} \times 1.319 = 0.8798 \;mol

×1.319=0.8798mol

V = \frac{nRT}{p} \\ V(H_2S) = \frac{0.8798 \times 0.08206 \times 463}{1.036}=32.26 \;LV=

nRT

V(H

S)=

1.036

0.8798×0.08206×463

=32.26L

Answer: 32.26 L

Explanation:

5 0

3 years ago