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velikii [3]
3 years ago
8

Burning one gallon of gasoline releases 20 pounds of CO2 into the air. Harry’s compact car can travel 25 miles on one gallon of

gas. If Harry drives 15,000 miles in a year, his vehicle will add pounds of CO2 to the atmosphere.
Chemistry
1 answer:
son4ous [18]3 years ago
5 0
There are several information's already given in the question. Based on those information's, the answer can be easily deduced.
Amount of gasoline required by Harry's car to travel 25 miles = 1 gallon
Then
amount of gasoline required
by Harry's car to travel 15000 miles = 15000/25
                                                         = 600 gallons
So 
Amount of CO2 released by burning 1 gallon of gasoline = 20 pounds
Then
Amount of CO2 released
by burning 600 gallon of gasoline = 600 * 20
                                                      = 12000 pounds
From the above deduction, it can be concluded that the amount of CO2 that will be added by Harry's car to the atmosphere is 12000 pounds.
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What is the pH of 0.000134 M solution of HCI?
mafiozo [28]

Answer: The pH will be 3.87

Explanation:

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

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

HCl\rightarrow H^++Cl^{-}

According to stoichiometry,

1 mole of HCl gives 1 mole of H^+

Thus 0.000134 moles of HCl gives =\frac{1}{1}\times 0.000134=0.0001342 moles of H^+

Putting in the values:

pH=-\log[0.000134]

pH=3.87

Thus the pH will be 3.87

5 0
3 years ago
Read 2 more answers
If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
3 years ago
The value of the entropy change for the process N₂ (g) + 3H₂ (g) --> 2NH₃ (g) is ________.
kodGreya [7K]

Answer:

negative

Explanation:

Entropy is a measure of the "disorder" in a system.

In this reaction, the amount of disorder decreases. This is because one gas molecule (NH₃) has more order than two gas molecules (N₂ and H₂). Therefore, the entropy change should be negative.

5 0
2 years ago
What is the National Suicide rate for Teens?
Butoxors [25]
I think it is 11% I read it on a article on msn.
4 0
4 years ago
Use Hess's Law to calculate the enthalpy change for the reaction
Marysya12 [62]

Answer:

ΔH = 125.94kJ

Explanation:

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:

1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ

2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ

-1/2 (1):

WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ

3/2 (2):

3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ

The sum of  last both reactions:

WO3(s) + 3H2(g) → W(s) + 3H2O(g)

ΔH = 842.7kJ -716.76kJ

<h3>ΔH = 125.94kJ </h3>
3 0
3 years ago
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