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Lera25 [3.4K]
3 years ago
13

a gas container in a thick walled balloon. when the pressure changes from __kPa to 114kpa, the volume changes from 0.654L to 1.3

2 L and the temperature changes from 596 K to 715 k
Chemistry
1 answer:
Hunter-Best [27]3 years ago
4 0

<span>To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At number of moles the value of PV/T is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:</span>

P1V1/T1 = P2V2/T2

P1 = P2V2T1/T2V1

P1 = (114)(1.32)(596)/(715)(.654)

P1 = 191.80 kPa

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Why do some element’s symbols not match their name? Example- iron=FE
Rina8888 [55]

Answer:

Because some of the symbols are from Latin

They are:

Mercury - Hg - Hydrargyrum

Potassium - K - Kalium

Silver - Ag - Argentum

Gold - Au - Aurum

Sodium - Na - Natrium

Tin - Sn - Stannum

Iron - Fe - Ferrum

Lead - Pb - Plumbum

Copper - Cu - Cuprum

Antimony - Sb - Stibium

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3 0
3 years ago
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adelina 88 [10]
<span>Only one parent and the same chromosome
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5 0
3 years ago
Ammonia is formed according to the reaction below. A chemist mixes 21 grams of nitrogen gas and 18 grams of hydrogen gas in a 2.
Rama09 [41]

Answer : The mass of hydrogen gas consumed will be, 4.5 grams

Explanation : Given,

Mass of N_2 = 21 g

Mass of H_2 = 18 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{21g}{28g/mole}=0.75moles

\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{18g}{2g/mole}=9moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given balanced reaction, we conclude that

As, 1 mole of N_2 react with 3 moles of H_2

So, 0.75 moles of N_2 react with 3\times 0.75=2.25 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent because it limits the formation of product.

The moles of hydrogen gas consumed = 2.25 mole

Now we have to calculate the mass of hydrogen gas consumed.

\text{Mass of }H_2=\text{Moles of }H_2\times \text{Molar mass of }H_2

\text{Mass of }H_2=(2.25mole)\times (2g/mole)=4.5g

Therefore, the mass of hydrogen gas consumed will be, 4.5 grams

5 0
4 years ago
To make lemonade, a recipe requires 4 lemons, 30 ounces of sugar, and 2 pints of water. From an average lemon one can squeeze ou
tresset_1 [31]

Answer:

She will need to add 5.6 cups of water

Explanation:

Hi there!

Marta has to dilute the citric acid to 5%.

The dilution factor will be 12%/ 5% = 2.4. Then, Marta will need to dilute the citric acid 2.4 times. If she has 4 cups of the solution, she will need to add water until she completes a volume of (4 cups ·2.4) 9.6 cups to reach the desired concentration.

Then, she will need to add 9.6 - 4 cups water = 5.6 cups of water

Another way to solve this is by using the fact that the mass of citric acid in the concentrated and diluted solution is the same. Then:

mass citric acid concentrated solution = mass citric acid in dilute solution

mass of citric acid = concentration · volume

Then:

initial concentration · volume = final concentration  · volume

12% · 4 cups = 5% · volume

volume = 12% · 4 cups / 5% = 9.6 cups

The final volume of the solution at 5% will be 9.6 cups. So Marta will need 9.6 cups - 4 cups = 5.6 cups water

3 0
4 years ago
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

  • Partial pressure of methane (pCH₄): 431 mmHg
  • Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0
3 years ago
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