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natita [175]
3 years ago
10

If 3.50 L of propane at 24°C and 62.2 kPa is completely burned in excess oxygen, how many moles of carbon dioxide will be produc

ed?

Chemistry
1 answer:
Eva8 [605]3 years ago
4 0
1) convert Celsius to Kelvin
24 + 273 = 297

2) convert kPa to atm
62.2 / 101.3 = 0.614

3) plug numbers into the ideal gas law formula

Answer: 0.088 mols

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Gamma rays

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Answer:

When the surface water is warm, the storm sucks up heat energy from the water, just like a straw sucks up a liquid. This creates moisture in the air. If wind conditions are right, the storm becomes a hurricane. This heat energy is the fuel for the storm.

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2 years ago
Crossing a Dd parent with another Dd parent results in
MrRissso [65]
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3 0
3 years ago
The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8
scZoUnD [109]

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane =\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M

The given balanced equilibrium reaction is,

                            cyclohexane  ⇔  methylcyclopentane

Initial conc.                 0.025 M           0

At eqm. conc.       (0.025-x)M       (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

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

0.12=\frac{(x)}{(0.025-x)}

By solving the term 'x', we get :

x =  0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

4 0
3 years ago
Gold is currently trading at very high price. Suppose that gold is selling for around $1860/ounce. How
Ugo [173]

Answer:

The answer is "3.81041978"

Explanation:

\to 1 \ OZ= 28.349523125 \ grams\\\\

              =28.349523125\times {1000} \ miligrams\\\\= 28349. 5231  \ miligrams\\

In 1860 = 28349.5231 \ miligrams\\

\to In \$ \ 1 = \frac{28349.5231}{1860} \ \ miligrams\\

             = 15.2416791 \ miligrams

\ In \  1 \ quarter =  \$ \ 0.25

\to \$ \ 0.25 =  15.2416791  \times 0.25 \  miligrams\\

              = 3.81041978

3 0
3 years ago
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