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HACTEHA [7]
1 year ago
10

8. A 220 mL sample of helium gas is in a cylinder with a movable piston at 105 kPa and 275K. The piston

Chemistry
1 answer:
Harrizon [31]1 year ago
8 0

The sample has a new pressure of 274kPa. If at 105 kPa and 275K, a 220 mL sample of helium gas is contained in a cylinder with a moving piston. The sample is pushed till it has a 95.0 mL volume and 310K .

The macroscopic characteristics of ideal gases are related by the ideal gas law (PV = nRT). A gas is considered to be perfect if its particles (a) do not interact with one another and (b) occupy no space (have no volume). Where P= pressure  V= volume and T = temperature.

From ideal gas equation

P₁V₁/T₁ =P₂V₂/T₂

105×220÷275 = P₂ ×95÷310

P₂= (105×220×310)÷(275×95)

P2= 7161000/26125

P2 = 274.105 kPa

Hence, the new pressure of helium gas is 274kPa

To know more about Ideas gas equation

brainly.com/question/28837405

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If Kc = 4.0×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 520 K , what is the value of Kp for this reaction at this temperature?
Flauer [41]

Here we have to get the K_{p} of the reaction at 520 K temperature.

The K_{p} of the reaction is 1.705 atm

We know the relation between K_{p} and K_{c} is K_{p}=K_{c}(RT)^{N}, where  K_{p} = The equilibrium constant of the reaction in terms of partial pressure, K_{c}  = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.

Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅

Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole

The given value of  K_{c} is 4.0×10⁻²

The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.

On plugging the values in the equation we get,

K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}

Or, K_{p} = 1.705 atm

Thus, the K_{p} of the reaction is 1.705 atm

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Which of the following group 7A elements is the most reactive ?
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The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)
cestrela7 [59]

Answer:

a. The limiting reactant is NaHCO_{3}

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}⇒3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

  • Na:  23
  • H: 1
  • C: 12
  • O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

  • NaHCO_{3} :23+1+12+16*3=84 g/mol
  • H_{3} C_{6} HO_{7} :1*3+12*6+1*5+16*7= 192 g/mol
  • CO_{2} :12+16*2= 44 g/mol
  • H_{2} O :1*2+16= 18 g/mol
  • Na_{3} C_{6} H_{5} O_{7} : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of NaHCO_{3} react with 1 mole of H_{3} C_{6} HO_{7}  Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:  

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of NaHCO_{3} react with 3 mole of CO_{2}. Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}

<u><em>grams of carbon dioxide=  0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

grams of citric acid=\frac{1.4 g * 192 g}{252 g}

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>

3 0
3 years ago
Chemistry help!
harkovskaia [24]

Answer:

Use the formula below

Explanation:

use the moles ratio for this by writing down the reaction and balancing the equation

4 0
2 years ago
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