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Volgvan
3 years ago
15

The enthalpy of combustion of hard coal averages -35 kJ/g, that of gasoline, 1.28 x 105 kJ/gal. How many kilograms of hard coal

provide the same amount of hat as is available from 1.0 gallon gasoline? Assume that the density of gasoline is 0.692 g/mL(the same s density of isooctane).
Chemistry
1 answer:
Ganezh [65]3 years ago
3 0

Answer:

3.657 kg

Explanation:

Given:

Enthalpy of combustion of hard coal = -35 kJ/g

Enthalpy of combustion of gasoline = 1.28 × 10⁵ kJ/gal

Density of gasoline = 0.692 g/mL

now,

The heat provide 1 gallon of gasoline provide =  1.28 × 10⁵ kJ

and,

heat provided by the 1 gram of coal  = 35 kJ

or

1 kJ of heat is provided by (1/35) gram of hard coal

therefore,

For 1.28 × 10⁵ kJ of heat, mass of hard coal =  1.28 × 10⁵ kJ × (1 / 35)

or

For 1.28 × 10⁵ kJ of heat, mass of hard coal = 3657.14 grams = 3.657 kg

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Which term describes a section of Earht's structure below moving continents as parts of large plates?
Snezhnost [94]

"Asthenosphere" describes a section of Earth's structure below moving continents as parts of large plates.

Answer: Option C

<u>Explanation: </u>

The asthenosphere denoted as the upper mantle of a very sticky, mechanically weak and flexible area in the earth. It is located below the surface of the lithosphere at depths of about 80 and 200 km (in terms of miles, 50 and 120). The boundary is commonly called the lithospheric LAB - asthenosphere.

The asthenosphere is nearly solid, though some of its regions could be molten below mid-ocean ridges for an instance. The lower asthenosphere boundary isn't well defined. The asthenosphere's thickness depends primarily on the temperature.

8 0
3 years ago
Consider the hypothetical reaction 3A + 4B → C + 2D Over an interval of 2.50 s the average rate of change of the concentration o
faust18 [17]

Answer:

Final [B] = 1.665 M

Explanation:

3A + 4B → C + 2D

Average rection rate = 3[A]/Δt = 4[B]/Δt = [C]/Δt = 2[D]/Δt

0.05600 M/s = 4 [B]/ 2.50 s

[B] = 0.035 M (concentration of B consumed)

Final [B] = initial [B] - consumed [B]

Final [B] = 1.700 M - 0.035 M

Final [B] = 1.665 M

3 0
2 years ago
The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins
saul85 [17]
The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>t_{1/2} = 2.5 min

We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{10 min}{2.5 min}
⇒ n=4<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>(1/2)^{n} = x
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<span>⇒x= \frac{1}{16}</span>
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3 years ago
A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance
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Answer:

W\geq 2.1x10^{-19}J

Explanation:

Due to Coulomb´s law electric force can be described by the formula F=K\frac{q_{1}.q_{2}}{r^{2}}, where K is the Coulomb´s constant (9x10^{9} N\frac{m^{2} }{C^{2} }), q_{1}= Charge 1 (Na+ in this case), q_{2} is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is W=W_{f} -W_{i}.

so we have W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]

Given that ri= 1.1nm= 1.1x10^{-9}m and rf= infinite distance

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