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Dimas [21]
2 years ago
14

Natural gas (CH4) has a molar mass of 16.0 g/mole. You started out the day with a tank containing 200.0 g of natural gas. At the

end of the day, your tank contains 30.0 g of natural gas. How many moles of gas did you use
Chemistry
1 answer:
hodyreva [135]2 years ago
3 0

Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas.  So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}

<u><em>amount of moles at the beginning= 12.5 moles</em></u>

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}

<u><em>amount of moles at the end= 1.875 moles</em></u>

The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.

moles used= amount of moles at the beginning - amount of moles at the end

moles used= 12.5 moles - 1.875 moles

<u><em>moles used= 10.625 moles</em></u>

<u><em /></u>

Finally, the moles of gas used are 10.625 moles.

Learn more about molar mass:

brainly.com/question/5216907

brainly.com/question/11209783

brainly.com/question/7132033

brainly.com/question/17249726

#SPJ1

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Question 2
Alenkinab [10]

Answer:

1.53 atm

Explanation:

From the question given above, the following data were obtained:

Volume = constant

Initial pressure (P₁) = stp = 1 atm

Initial temperature (T₁) = 273 K

Final temperature (T₂) = 144 °C = 144 °C + 273 = 417 K

Final pressure (P₂) =?

Since the volume is constant, the final pressure can be obtained as follow:

P₁ / T₁ = P₂ / T₂

1 / 273 = P₂ / 417

Cross multiply

273 × P₂ = 417

Divide both side by 273

P₂ = 417 / 273

P₂ = 1.53 atm

Therefore, the final pressure (i.e the pressure inside the hot water bottle) is 1.53 atm.

8 0
3 years ago
If 16.5g of c6h14o2 are reacted vwith .499 mol of o2. How many moles of co2 should be produced
Fed [463]

Answer:

moles of carbon dioxide produced are 410.9 mol.

Explanation:

Given data:

Mass of C₆H₁₄O₂ = 16.5 g

Moles of O₂ = 499 mol

Moles of CO₂ = ?

First of all we will write the balance chemical equation.

2C₆H₁₄O₂  +  17O₂  →   14CO₂  +  12H₂O

moles of C₆H₁₄O₂  = mass × molar mass

moles of C₆H₁₄O₂ =  16.5 g × 118 g/mol

moles of C₆H₁₄O₂ = 1947 mol

Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.

                 O₂      :     CO₂

                 17       :      14

                499     :      14/17× 499 = 410.9 moles

          C₆H₁₄O₂   :   CO₂

                    2    :      14

                 1947 :     14/2× 1947 =  13629 moles

Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.

5 0
4 years ago
Read 2 more answers
How many moles in 110g of nahco3?
kenny6666 [7]
Hope this helps you.

5 0
3 years ago
Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib
Dvinal [7]

<u>Answer:</u> The value of K_c for the final reaction is 7.16\times 10^{25}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1

<u>Equation 2:</u>  HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2

The net equation follows:

S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=\frac{1}{K_1}\times \frac{1}{K_2}

We are given:

K_1=9.57\times 10^{-8}

K_2=1.46\times 10^{-19}

Putting values in above equation, we get:

K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}

Hence, the value of K_c for the final reaction is 7.16\times 10^{25}

5 0
3 years ago
How does the law of definite proportions apply to hydrates?
vladimir1956 [14]
The law of definite proportions would state that a hydrate always contain exactly the same proportion of salt and water by mass.
strictly speaking, the law of definite proportion states that a compound always 
contains exactly the same proportion of elements by mass.
But the law is often applied to groupings of elements in compound.
Hydrates are salt that have a certain amount of  water asa part of their structure.
The water is chemically combined with the compound in a definite ratio.
4 0
3 years ago
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