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avanturin [10]
3 years ago
6

Consider the synthesis of ammonia 3 H2+ N2 à 2 NH3 If a student were to react 38.5 g of nitrogen gas, how many moles of ammonia

would be produced?
Chemistry
1 answer:
Wittaler [7]3 years ago
4 0

Answer:

2.75 mol

Explanation:

Given data:

Mass of Nitrogen = 38.5 g

Moles of ammonia produced = ?

Solution:

Chemical  equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

           N₂            :            NH₃

            1              :             2

           1.375       :           2×1.375 = 2.75 mol

Thus 2.75 moles of ammonia  are produced from 38.5 g of nitrogen.

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Find the pH of a 0.00476M HCI solution​
Alex777 [14]

Ans:2.3

Explanation:

pH = -log[H+(aq)]

= -log[0.00476]

=2.3

As HCl solution is monobasic. Each molecule ionizes in water to form ONE hydrogen ion. So the H+(aq) is equal to its molarity.

5 0
3 years ago
Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2.
ludmilkaskok [199]

Answer:

Percentage yield =  25%

Explanation:

Given data:

Moles of C₈H₁₈ = 4 mol

Moles of O₂ = 4 mol

Actual yield of CO₂ = 28.16 g

Percentage yield = ?

Solution:

Chemical equation:

2C₈H₁₈ + 25 O₂ → 16CO₂ + 18H₂O

Now we will compare the moles of CO₂ with O₂ and C₈H₁₈.

                 C₈H₁₈     :      CO₂

                     2        :       16

                     4        :       16/2×4=32 mol

                  O₂         :      CO₂

                    25        :      16

                     4         :       16/25×4=2.56 mol

Mass of CO₂:

Mass = number of moles× molar mass

Mass = 2.56 mol ×44 g/mol

Mass = 112.64 g

Percentage yield:

Percentage yield = Actual yield /theoretical yield × 100

Percentage yield = 28.16 g / 112.64 g× 100

Percentage yield = 0.25× 100

Percentage yield =  25%

5 0
3 years ago
if you drop a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius, what best descr
Ymorist [56]
As soon as the metal gets into the water a temperature transfer will start between both of them resulting in  

1. Metal will cool down. 
2. Some water will get evaporated.
3. Temperature of water will rise.
4. As the metal gets into the water, a sound like chissssss will be heard.

Hope it helps.
Don't be shy and say thx.<span />
8 0
3 years ago
How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(C=0.895 Jig Cy from 22.5°C to 55 0
dsp73

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

q=mc\Delta T\\\\q=mc(T_2-T_1)

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum = 0.895J/g^oC

\Delta T = change in temperature

T_1 = initial temperature = 22.5^oC

T_2 = final temperature = 55.0^oC

Putting values in above equation, we get:

q=223g\times 0.895J/g^oC\times (55.0-22.5)^oC

q=6486.5125J

Therefore, the heat energy needed would be, 6486.5125 J

5 0
3 years ago
Molecules are likely to dissolve in nonpolar solvents.
masha68 [24]
The answer to this is yes they are likely, true.
4 0
3 years ago
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