Ask question

Ask question

All categories

2 years ago

14

Considering the combustion reaction of propane again, what mass of propane (C3Hg) is necessary to react with 4.53 g of oxygen? C

3H8 +5023CO₂ + 4H₂O 0.0256 g C3H8 1.13 g C3H8 1.25 g C3H8 2.50 g C3H8

1 answer:

zimovet [89]2 years ago

7 0

Answer:

1.25 g C3H8

Explanation:

edge lol

You might be interested in

The ph of 0.015 m hno2 (nitrous acid) aqueous solution was measured to be 2.63. what is the value of pka of nitrous acid?

Licemer1 [7]

Nitrous acid<span> dissociates as follows:
</span>
HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq)

According to the equation, an acid constant has the following form:

Ka = [H⁺] × [NO₂⁻ ] / [HNO₂]

From pH, we can calculate the concentration of H⁺ and NO₂⁻:

[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]

Now, the acid constant can be calculated:

Ka = 0.00234 x 0.00234 / 0.015 = 3.66 x 10⁻⁴

And finally,

pKa = -log Ka = 3.44

7 0

2 years ago

PLEASE ANSWER ASAP!

uranmaximum [27]

Not for sure but i would say the its B rate please

6 0

2 years ago

30. In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white

grigory [225]

Answer: 2Mg + O2 》 2MgO

3Mg + N2 》Mg3N2

Explanation:

6 0

2 years ago

Chose the molecular formula that represents an organic molecule.

g100num [7]

Answer:

(D)C 6H 12O 6

Explanation:

7 0

2 years ago

The bomb calorimeter in Exercise 102 is filled with 987g water. The initial temperature of the calorimeter contents is 23.32. A

Sholpan [36]

Answer:

25.907°C

Explanation:

In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C

The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.

Thus:

$-q_{combust} = q_{water} + q_{calori}$

$q_{combust}$ = heat of combustion of benzoic acid

$q_{water}$ = heat energy released to water

$q_{calori}$ = heat energy released to the calorimeter

Therefore,

$-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})$

1.056*26.42 = [0.987*4.18 + 6.66]( $T_{f}$ - 23.32)

27.8995 = [4.12566+6.660]( $T_{f}$ - 23.32)

( $T_{f}$ - 23.32) = 27.8995/10.7857 = 2.587

$T_{f}$ = 23.32 + 2.587 = 25.907°C

4 0

3 years ago