Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
Answer:
The correct answer is - 29.45 / 100 x 25.6 = 7.5392 grams
Explanation:
It is given in the question that in 100 gms of CaSO4 there are 29.45 grams of Ca present and there is 25.6 gram of total CaSO4 sample present, So, to calculate the exact value of calcium in this given sample is:
mass of Ca = total amount of sample*percentage of calcium in sample /100
M of Ca =25.6*29.45/100
M of Ca = 7.5392 grams
Thus, the correct procedure is given by 29.45 / 100 x 25.6 = 7.5392 grams
Answer:
a mixture of two these
Explanation:
The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.
n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.
Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.
Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.
Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
To learn more about pH here
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