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Degger [83]
3 years ago
14

How many kilograms of gasoline fill a 12.0-gal gas tank

Chemistry
1 answer:
Sergeu [11.5K]3 years ago
8 0

Answer:

29.98kg

Explanation:

12.0 gallons * (3.78541178 liters/gallon) * (1000 mL/liter) * (0.66 g/mL) * (1 kg/1000 g) = 29.98 kg

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Why do substances with weak intermolecular forces form solids at low melting points?
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To make any substance solid, molecules must come very close to each other. Substances with weak intermolecular forces have weak bonding. Hence to make their molecules come close to each other, we must provide low temperature. 
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3 0
3 years ago
What do a mole of magnesium (Mg) and a mole of iron (Fe) have in common? their number of atoms their number of molecules their n
-Dominant- [34]
It has the same number of atoms
3 0
3 years ago
If 71.5 moles of an ideal gas is at 5.03 atm at 6.80 °C, what is the volume of the gas?
meriva
Use the clapeyron equation:

T in kelvin : 6.80 + 273 => 279.8 K

R = 0.082 

n = 71.5 moles

P = 5.03 atm

Therefore:

P x V = n x R x T

5.03 x V = 71.5 x 0.082 x 279.8

5.03 x V = 1640.4674

V = 1640.4674 / 5.03

V = 326.13 L

hope ths helps!
5 0
3 years ago
Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the
Cloud [144]

Answer : The equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

Solution : Given,

Equilibrium constant, K_c=1.8\times 10^{-5}

Initial concentration of HC_2H_3O_2 = 0.260 m

Let, the 'x' mol/L of H_3O^+ are formed and at same time 'x' mol/L of HC_2H_3O_2 are also formed.

The equilibrium reaction is,

                  HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)

Initially                0.260 m                       0                 0

At equilibrium    (0.260 - x)                     x                 x

The expression for equilibrium constant for a given reaction is,

K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

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

1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}

By rearranging the terms, we get the value of 'x'.

x=2.154\times 10^{-3}m

Therefore, the equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

4 0
3 years ago
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