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When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.

d1i1m1o1n [39]

Answer:

48%

Explanation:

Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:

P1T2 = P2T1

Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:

P1 = 10.0atm

T2 = 1420K

P2 = ?

T1 = 730K

P2 = 10.0atm*1420K / 730K

P2 = 19.45 atm

The CO2 reacts as follows:

2CO2 → 2CO+ O2

Where 2 moles of gas react producing 3 moles of gas

Assuming the 100% of CO2 react, the pressure will be:

19.45atm * (3mol / 2mol) = 29.175atm

As the pressure rises just to 24.1atm the moles that react are:

24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present

The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:

0.48mol / 1mol * 100 = 48%

8 0

3 years ago

In which group of teh modern periodic table are there very reactive metals and very reactive non metals?

Rufina [12.5K]

Metals :-

Group 1A - Alkali metals ( highly reactive metals)

Non-metals :-

Group 17 - Halogens ( highly reactive non-metals )

5 0

3 years ago

In this simulation, ___________________ will be used to stimulate the axon.

inysia [295]

Voltage is known to be used as a requirement for stimulating an axon. The axon conducts electrical signals away from the nerve cell. The main function of the axons is that it is specifically used to transmit information, in a form of electrical impulse, to the other parts of the body.

5 0

3 years ago

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

slava [35]

Answer: The theoretical yield of iron(III) sulfate is 26.6 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago

When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu

IgorLugansk [536]

When connectors are marked with a combination of metals, it can be used as a connector of one of the metals or an alloy of the two metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and Cu is copper, therefore the answer is:

Yes, it is suitable for use with copper, copper-clad aluminum, and aluminum conductors.

6 0

3 years ago