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NISA [10]
3 years ago
6

Which of the following is not a type of fault?

Chemistry
1 answer:
bekas [8.4K]3 years ago
8 0

normal is the answer


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How many calories are required to raise the temperature of 75g of water from 20 ˚C to 50˚ C?
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The answer to the question is letter b
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Can a combination reaction be a redox reaction ?
saul85 [17]

Answer:

Yes A redox reaction can be a combination reaction.

Two elements are mixed in a combination reaction to produce a single product.

Explanation:

Example: water formula

2H2 + O2 → 2H2O

Oxygen is reduced in this reaction when electrons are transferred from hydrogen to oxygen and hydrogen is also oxidized since oxygen from hydrogen accepts electrons. Oxygen is the oxidizing agent and hydrogen is the reducing agent.

2H2 + O2 → 2H2O

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A typical phase change diagram is shown below. Which of the numbered arrows represent changes where the kinetic energy of the pa
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Do you have a diagram?

7 0
4 years ago
A tank at is filled with of sulfur tetrafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal g
dangina [55]

The question is incomplete, the complete question is:

A 7.00 L tank at 21.4^oC is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

<u>Answer:</u> The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

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

  • <u>For sulfur hexafluoride:</u>

Given mass of sulfur hexafluoride = 5.43 g

Molar mass of sulfur hexafluoride = 146.06 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol

  • <u>For sulfur tetrafluoride:</u>

Given mass of sulfur tetrafluoride = 14.2 g

Molar mass of sulfur tetrafluoride = 108.07 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol

Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

\chi_A=\frac{n_A}{n_A+n_B} .....(2)

where n is the number of moles

Putting values in equation 2, we get:

\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221

\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779

Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

7 0
3 years ago
CaCO3(s) ⇄ CaO(s) + CO2(g) 0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K.
Montano1993 [528]

Answer:

The final mass of CaCO3 is 10.68 grams

Explanation:

Step 1: Data given

Number of moles CaCO3 = 0.100 moles

Number of moles CaO = 0.100 moles

Volume = 10.0 L

When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.250 atm of CO2 is added, while keeping the temperature constant

Step 2: The balanced equation

CaCO3(s) <==> CaO(s) + CO2(g)

Step 3: Calculate moles of CO2

n = PV/RT

⇒n = the initial number of moles CO2 = TO BE DETERMINED

⇒P = the pressure of CO2 at theequilibrium = 0.220 atm

⇒V = the volume of the container = 7.0 L

⇒R = the gas constant = 0.08206 L*atm / mol * K

⇒T = the temperature = 385 K

n = 0.220*7.0/(0.08206*385) = 0.0487 (mol)

this is the amount of CaCO3 which has been converted to CaO before pumping-in additional 0.225 atm CO2(g).

Step 4: Calculate moles CaCO3

After adding additional 0.250 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm.  All this additional CO2 would completely convert to CaCO3:

n = PV/RT = 0.250*7.0/(0.08206*385) = 0.0554 moles

The total CaCO3 after equilibrium is reestablished is:

0.100 - 0.0487+ 0.0554 = 0.1067 mol

Step 5: Calculate mass CaCO3

Mass CaCO3 = 0.1067 moles * 100.09 g/mol

Mass CaCO3 = 10.68 grams

The final mass of CaCO3 is 10.68 grams

8 0
3 years ago
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