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givi [52]
3 years ago
5

What is the pressure of the ideal gas when the volume is decreased to 4.0 L and the temperature is increased to 265 K?

Chemistry
1 answer:
jonny [76]3 years ago
8 0
It is an ideal gas therefore we can use the ideal gas equation to solve the problem. The ideal gas equation is expressed as PV = nRT. First, we solve the amount of the gas in moles using the said equation and the first conditions.

(2.0 atm) (5.0 x 10^3 cm^3) = n (82.0575 atm.cm^3/mol.K)(215 K)
n=0.5668 mol

Using the second conditions given, we obtain the new pressure.
P (4.0 x 10^3) = 0.5668 x <span>82.0575 x 265
P= 3.08 atm</span>
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Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338
Elden [556K]

<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For KCl:</u>

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

  • <u>For magnesium chloride:</u>

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = (2\times 0.224)=0.448mol

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

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Calculate the molar mass of iron(III) sulfide. Round your answer to the nearest 0.1 and include units to receive credit.
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Answer:

207.89g

Explanation:

The formula of the compound is:

        Fe₂S₃  

Find the molar mass of the compound;

  Atomic mass of Fe  = 55.845g/mol

                              S  = 32.065g/mol

Now;

  Molar mass of  Fe₂S₃   = 2 (55.845) + 3 (32.065)

                                         = 207.89g

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The elements in Group 2 (beryllium, magnesium, calcium, strontium, barium, and radium) are called the alkaline earth metals (see Figure below). These elements have two valence electrons, both of which reside in the outermost s sublevel. The general electron configuration of all alkaline earth metals is ns

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