Ask question

Ask question

All categories

3 years ago

5

For the complete combustion of 1.000 mole of methane gas at 298 K and 1 atm pressure, ΔH° = -890.4 kJ/mol. What will be the amou

nt of heat released when 1.97 g of methane is combusted under these conditions?

1 answer:

anastassius [24]3 years ago

3 0

Oh hooray a practice problem:

That might be wrong

You might be interested in

What did the professer use a catalyst for the flubber experiment

Anettt [7]

In the 1997 movie called, Flubber, the professor used an organic catalyst and a little touch of electricity to make flubber. The organic substance is not known.
Thank you for posting your question here at brainly. I hope the answer helps.

7 0

2 years ago

How many liters of water vapor are in 36.21 g?

Elden [556K]

Answer:

45.02 L.

Explanation:

Firstly, we need to calculate the no. of moles of water vapor.

n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.

We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.

<em><u>Using cross multiplication:</u></em>

1.0 mole of CO occupies → 22.4 L.

2.01 mole of CO occupies → ??? L.

∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.

4 0

2 years ago

For an insoluble compound, how might you prevent formation of the precipitate

anastassius [24]

<span>protect it by covering it by lyophilic sol.</span>

8 0

2 years ago

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°

miskamm [114]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H_1=157kJ$ ( × 4)

(2) $P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)$ $\Delta H_2=-1207kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

6 0

2 years ago

Match the boxes helppppppp

lesantik [10]

Answer:

?

Explanation:

3 0

2 years ago