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notka56 [123]
1 year ago
5

2.00 g of an unknown gas at STP fills a 500. mL flask. What is the molar mass of the gas?

Chemistry
1 answer:
otez555 [7]1 year ago
4 0

Answer:

100g/mol

Explanation:

Given parameters:

Mass of unknown gas  = 2g

Volume of gas in flask  = 500mL  = 0.5dm³

Unknown:

Molar mass of gas = ?

Solution:

Since we know the gas is at STP;  

        1 mole of substance occupies 22.4dm³ of space at STP

    Therefore,

            0.5dm³ will have  0.02mole at STP

                     

Now;

   Number of moles  = \frac{mass}{molar mass}  

      Molar mass  = \frac{mass}{number of moles}   = \frac{2}{0.02}   = 100g/mol

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If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T
zhuklara [117]

Answer: The actual yield of B_2O_3 is 60.0 g

Explanation:-

The balanced chemical reaction :

B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)

Mass of B_2H_6 =Density\times Volume=1.131g/ml\times 36.9ml=41.7g

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles

According to stoichiometry:

1 mole of B_2H_6 gives = 1 mole of B_2O_3

1.51 moles of B_2H_6 gives =\frac{1}{1}\times 1.51=1.51 moles of B_2O_3

Theoretical yield of B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g

Percent yield of B_2O_3= 75.7\%

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

75.7\%=\frac{\text{Actual yield}}{79.3}\times 100

{\text{Actual yield}}=60.0g

Thus the actual yield of B_2O_3 is 60.0 g

7 0
2 years ago
1. A crane lifts a 75kg mass a height of 8 m. Calculate the gravitational potential energy
koban [17]

Answer:

GP.E = 5880 j

Explanation:

Given data:

Mass = 75 kg

height = 8 m

Potential energy = ?

Solution:

The formula for gravitational potential energy is

GPE = mgh

m = mass in kilogram

g = acceleration due to gravity

h = height in meter above the ground

Formula:

GP.E = mgh

Now we will put the values in formula.

g = 9.8 m/s²

GP.E = 75 Kg × 9.8 m/s²× 8 m

GP.E = 5880 Kg.m²/s²

Kg.m²/s² = j

GP.E = 5880 j

6 0
1 year ago
1) How many aluminum atoms are there in 3.50 grams of Al2O3?
drek231 [11]

Answer:

Aluminium atoms =  4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

Step 1: Data given

Mass of Al2O3 = 3.50 grams

Molar mass of Al2O3 = 101.96 g/mol

Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

Moles Al2O3 = 0.0343 moles

Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al

Step 4: Calculate aluminium atoms

Aluminium atoms =  moles aluminium * Number of Avogadro

Aluminium atoms =  0.0686 * 6.022 * 10^23

Aluminium atoms =  4.13 *10^22 aluminium atoms

The correct answer is E

3 0
2 years ago
Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of
Alecsey [184]

Answer:

Without dark matter galaxies would loose an extreme amount of gas required to create stars.

Without dark matter the universe wont have as many galaxies clumped together forming larger versions of those galaxies. This would cause a change in the structure of the "skeleton" of the web.

(Hope this can help, I didn't do exactly as it is said to because that is your job)

:)

Explanation:

Forbes gives somewhat of an explanation if you are curious.

(Ethan Siegal, "The Universe Would Be Very Different Without Dark Matter", Forbes)

3 0
1 year ago
Both black and white road surfaces radiate energy. at midnight on a starry night the warmer road surface is the
luda_lava [24]
The black road because during the day it absorbed more radiation than the with one
5 0
1 year ago
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