Answer:
Option A
Explanation:
Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.
Answer:
The number ratio is 4:7
Explanation:
Step 1: Data given
Compound 1 has 50.48 % oxygen
Compound 2 has 36.81 % oxygen
Molar mass oxygen = 16 g/mol
Molar mass manganese = 54.94 g/mol
Step 2: Calculate % manganes
Compound 1: 100 - 50.48 = 49.52 %
Compound 2: 100 - 36.81 = 63.19 %
Step 3: Calculate mass
Suppose mass of compounds = 100 grams
Compound 1:
50.48 % O = 50.48 grams
49.52 % Mn = 49.52 grams
Compound 2:
36.81 % O = 36.81 grams
63.19 % Mn = 63.19 grams
Step 4: Calculate moles
Compound 1
Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles
Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles
Compound 2
Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles
Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles
Step 5: calculate mol ratio
We will divide by the smallest amount of moles
Compound 1
O: 3.155/0.9013 = 3.5
Mn: 0.9013 / 0.9013 = 1
Mn2O7
Compound 2
O: 2.301 / 1.150 = 2
Mn: 1.150 / 1.150 = 1
MnO2
The number ratio is 2:3.5 or 4:7
The artificial fixation of nitrogen (N2) has enormous energy, environmental, and societal impact, the most important of which is the synthesis of ammonia (NH3) for fertilizers that help support nearly half of the world's population.
<h3>Artificial fixation of nitrogen</h3>
a) The equilibrium constant expression is Kp=PCH4 PH2 OP CO×PH 23.
(b) (i) As the pressure increases, the equilibrium will shift to the left so that less number of moles are produced.
(ii) For an exothermic reaction, with the increase in temperature, the equilibrium will shift in the backward direction.
(iii) When a catalyst is used, the equilibrium is not disturbed. The equilibrium is quickly attained
To learn more about equilibrium constant visit the link
brainly.com/question/10038290
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Answer:
0.106 mol (3s.f.)
Explanation:
To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.
Ar of C= 12, Ar of H= 1, Mr of O= 16
These Ar values can be found on the periodic table.
Mr of glucose= 6(12)+ 12(1) + 6(16)= 180
Moles of glucose
= mass ÷ mr
= 19.1 ÷ 180
= 0.106 mol (3 s.f.)
