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Llana [10]
3 years ago
8

A thick, oily liquid found deep in the earth that is refined and used as a source for gasoline is called _____________?

Chemistry
1 answer:
grandymaker [24]3 years ago
6 0

It would be A, Petroleum.

Hope this helped!

-William

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A steel container with a movable piston contains 2.00 g of helium which was held at a constant temperature of 25 °C. Additional
shtirl [24]

Answer: D) 1.00 g

Explanation:

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = 2.00 L

V_2 = final volume of gas = 3.00 L

n_1 = initial moles of gas  =\frac{\text {Given mass of helium}}{\text {molar mass of helium}}=\frac{2.00g}{4g/mol}=0.500mol

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{2.00L}{3.00L}=\frac{0.500mol}{n_2}

n_2=0.75mole

Mass of helium =moles\times {\text {molar mass}}=0.75\times 4=3.00g

Thus mass of helium added = (3.00-2.00) g = 1.00 g

4 0
3 years ago
Can you help me please
irina [24]

Answer:

Concentrated solutions

7 0
3 years ago
Hydrazine (N₂H₄), a rocket fuel, reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the e
Vinvika [58]

At STP one mol weighs 22.4L

Moles of O_2

  • 4.2/22.4
  • 2.1/11.2
  • 0.19mol

1 mol.O_2 can create 2mol water

moles of water

  • 2(0.19)
  • 0.38mol

Volume of water

  • 0.38(22.4)
  • 8.512L
8 0
2 years ago
Read 2 more answers
Which acid has the largest ka: hclo3(aq), hbro3(aq) or hio3(aq)?
ryzh [129]
Greater the Ka value greater is the acid strength. Among these three acids HClO3 is the strongest acid, with greatest Ka and lowest pKa value (-1), Then comes HBrO3, its the second most strongest acid among the three, its Pka value is 0.7, higher than HClO3 but smaller than HIO3 (i.e. 0.77) which the weakest acid among the three.

3 0
4 years ago
How much of 45 grams of pu 234 remains after 27 hours if it’s half life is 4.98 hours
nadya68 [22]

Answer:

1.07 g

Explanation:

Half-life of Pu-234 = 4.98 hours

Initially present = 45 g

mass remains after 27 hours = ?

Solution:

Formula

         mass remains = 1/ 2ⁿ (original mass) ……… (1)

Where “n” is the number of half lives

To find "n" for 27 hours

                          n = time passed / half-life . . . . . . . .(2)

put values in equation 2

                          n = 27 hr / 4.98 hr

                          n = 5.4

Mass after 27 hr

Put values in equation 1

          mass remains = 1/ 2ⁿ (original mass)

          mass remains = 1/ 2^5.4 (45 g)

          mass remains = 1/ 42.2 (45 g)

          mass remains =  0.0237 x 45 g

          mass remains =  1.07 g

6 0
4 years ago
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