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7nadin3 [17]
2 years ago
15

A 8.5-liter sample of a gas at 2.0 atm and 300.0 K has 1.2 moles of the gas. If 0.65 mole of the gas is added to the sample at t

he same temperature and pressure, what is the final volume of the gas?
13 liters
14 liters
18 liters
21 liters
Chemistry
2 answers:
hodyreva [135]2 years ago
7 0

Did you take the test? what was the answer Im stuck on this one too

choli [55]2 years ago
3 0

Answer: 13 liters

Explanation:

Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

V\propto n   (At constant temperature and pressure)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1 = initial volume of gas = 8.5 L

V_2 = final volume of gas = ?

n_1 = initial moles of gas =  1.2 moles

n_2 = final moles of gas = 1.2 +0.65 = 1.85 moles

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{8.5L}{1.2}=\frac{V_2}{1.85}

V_2=13L

Therefore, the final volume of the gas will be 13 Liters.

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Draw a structure for the product of nucleophilic substitution obtained on solvolysis of tert−butyl bromide in methanol, and arra
Elena-2011 [213]

Answer:

2-methoxy-2-methylpropane

Explanation:

The first step for this reaction is the carbocation formation. In this step, a tertiary carbocation is formed. Also, we will have a good leaving group so bromide will be formed. Then the methanol acts as a nucleophile and attacks the carbocation. Next, a positive charge is generated upon the oxygen, this charge can be removed when the hydrogen leaves the molecule as H^+. (See figure)

6 0
2 years ago
Calculate the molarity of a solution that contains 70.0 g of H2SO4 in 280. mL of solution.
maw [93]
First convert grams to moles:
70.0g *(mole/98.079) = 0.7137mole
Remember that molarity is moles per liter:
0.7137mole *(1/280mL) *(1000mL/L) = 25.5M
8 0
1 year ago
There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 12.01%. Now that you have the contribution
mestny [16]

Answer: The average atomic mass of X is 16.53

Explanation:

Mass of isotope X-19  = 2.282  

% abundance of isotope X-19 = 12.01% = \frac{12.01}{100}=0.1201

Mass of isotope X-21 = 18.48

% abundance of isotope X-21 = (100-12.01)% = \frac{100-12.01}{100}=0.8799

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(2.282\times 0.1201)+(18.48\times 0.8799)]

A=16.53

Therefore, the average atomic mass of X is 16.53

4 0
2 years ago
How many grams of water can be cooled from 41 ∘c to 19 ∘c by the evaporation of 62 g of water? (the heat of vaporization of wate
Vsevolod [243]
62 g of water are vaporized and the energy required is 2.4 kJ/g

So 62g x 2.4 kJ/g = 148.8 kJ or 148,800 Joules 

Q = mCΔT
Q is energy in joules, m is mass of water, C is the specific heat, delta T is change in temp

148,800 = m(4.18)(41 - 19) = 1618g or 1.6 kg of water

8 0
1 year ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
2 years ago
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