A summary of the Law of multiple proportions is that if A and B form more than one compound, and B1 is the amount of element B which reacts with a fixed mass of A in compound 1, and B2 is the amount of B which reacts with the same fixed mass of B to form compound 2, then the ratio B1:B2 will be small whole numbers.
This law is rather simplistic, and given the range of compounds known today the definition of 'small' is now rather large... but, to answer the question:
in compound one 1.14133g of B reacts with 1g of A. (1.14133=53.3/46.7)
We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, 
moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
decameters - meters: multiply by 10
meters to meters: multiply by 1
centimeters to meters: divide by 100
millimeters to meters: divide by 1000
For the rows at the bottom:
hectometer row: 100, multiply by 100, 4500
decameter row: 10, multiply by 10, 450
meter row: 1, multiply by 1, 45
decimeter row: 0.1, divide by 10, 4.5
centimeter row: 0.01, divide by 100, 0.45
im guessing theres a millimeter row at the bottom:
millimeter row: 0.001, divide by 1000, 0.045
hope this helps!
Answer:
See explanation
Explanation:
a) The equation of the reaction is;
2Na + Cl2 ------>2NaCl
Number of moles of sodium = 10g/23 g/mol = 0.43 moles
If 2 moles of sodium reacts with 1 mole of Cl2
0.43 moles reacts with 0.43 * 1/2 = 0.215 moles of Cl2
Mass = 0.215 moles of Cl2 *71 g/mol = 15.265 g
b) Equation of the reaction;
HgO -> Hg + O2
1.252 moles of HgO 1.252/32 gmol = 0.039 moles
1 mole of HgO yields 1 mole of oxygen hence
0.039 moles of HgO yields 0.039 moles of oxygen
Mass of oxygen = 0.039 moles * 32 g/mol = 1.248 g
c) Equation of the reaction;
2NaNO3 -----> 2NaNO2 + O2
Number of moles of 128 g of oxygen = 128g/32 g/mol = 4 moles
2 moles of NaNO3 yields 1 mole of oxygen
x moles of NaNO3 yields 4 moles of oxygen
x = 8 moles of NaNo3
Mass of NaNO3 = 8 * 85 g/mol = 680 g of NaNo3
Answer:
is the value of the equilibrium constant at this temperature.
Explanation:
Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as 
Partial pressures at equilibrium:
The equilibrium constant in terms of pressures is given as:
is the value of the equilibrium constant at this temperature.
