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True [87]
3 years ago
13

How many U.S. gallons are there in a cubic mile? The total proven oil reserves of the U.S. are roughly 30 x 10°bbl. How many cub

ic miles is this?
Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
4 0

Answer:

1 cubic mile = 1.101 * 10^12 US gallons

1 US bbl oil = 42 US gallons = 3.8143*10^ -11 cubic miles

Explanation:

The number of the exponent of the oil reserve is not very well shown in the question so, I provide you the conversion of bbl oil into cubic mile, the only thing you have to do is multiply the number of bbls of the reserve for the conversion in cubic miles and you'll have the answer.

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Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
3 years ago
The density of liquid mercury is 13.6 g/mL. What is its density in units of ? (2.54 cm = 1 in., 2.205 lb = 1 kg)
nalin [4]

Correct question

The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in​3​? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)

Answer:

\rho0.4916\ lb/in^3

Explanation:

Given that;-

The density = 13.6 g/mL

Also, 1 kg = 2.205 lb

1 kg = 1000 g

So, 1000 g = 2.205 lb

1 g = 0.002205 lb

Also,

1 in = 2.54 cm

1 in³ = 16.39 cm³

1 cm³ = 1 mL

So,  1 in³ = 16.39 mL

1 mL = 0.061 in³

The expression for the calculation of density is shown below as:-

\rho=\frac{m}{V}

Thus,

\rho=\frac{13.6\ g}{1\ mL}=\frac{13.6\times 0.002205\ lb}{0.061\ in^3}=0.4916\ lb/in^3

7 0
3 years ago
Which of the following is insoluble in water?
bija089 [108]

Answer:

I think Is A

Explanation:

5 0
3 years ago
A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will b
Angelina_Jolie [31]

Answer:

50.4 % of the water will be vapor

Explanation:

<u>Step 1:</u> Data given

Mass of water = 1.68 grams

volume of the flask = 5.3 L

Temperature = 65°C

Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

<u>Step 2:</u> Calculate moles of H2O

p*V=n*R*T

⇒ p = the pressure of water = 0.2467 atm

⇒ V = the volume of the flask = 5.3 L

⇒ n = moles of water

⇒ R = gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = 65°C = 338 Kelvin

n = (p*V)/(R*T)

n = (0.2467 * 5.3) /(0.08206* 338)

n = 0.047 moles

<u>Step 3:</u> Calculate mass of water

Mass of water = moles of water * molar mass of water

Mass of water = 0.047 moles *18.02 g/mol

Mass of water = 0.84694 grams

<u>Step 4:</u> Calculate the percent of water vaporized

% = (0.84694 grams/1.68 grams) *100%

% = 50.4%

50.4 % of the water will be vapor

5 0
3 years ago
A student was given a stock solution with a concentration of 3.61 g/mL and asked to perform
vovangra [49]

Answer:

The concentration of the dilute sample will be 0.361 g/ml

Explanation:

If a solution is diluted into 1:10 ratio then the amount of solute of that solution will be decreased by 10 times.

      The initial concentration of the stock solution was 3.61g/ml but when the solution is diluted in 1:10 ratio the solute concentration is also decreased by 10 times.SO at present the solute concentration becomes 3.61/10=0.361 g/ml.

8 0
3 years ago
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