In order for you to calculate
for the mass of ammonium carbonate, you need to know the molar mass of it and
the nitrogen atoms in the compound. Ammonium carbonate has a molar mass of
96.08 grams per mole. There are two nitrogen atoms in ammonium carbonate which
is equal to 28.02 grams per mole. Divide the molar mass of nitrogen to the
ammonium carbonate, 28.02/96.08 x 100, we get 29.16wt% nitrogen.
0.000169 mol/g citric acid
Answer:
Sulfur dioxide is a covalent bond because we have two non-metals (Sulfur and Oxygen).
(<em>Sulfur dioxide formula)</em>
It's called Sulfur dioxide but it can be called Sulfurous anhydride and Sulfur (lV) oxide.
Sulfur dioxide can have different reactions:
-
In the presence of oxygen oxidation of sulfur dioxide to sulfur trioxide occurs and the formula is:

The equal is because it's a reversible equation. This reaction is spontaneous.
- The sulfur trioxide reacting with water produces sulfuric acid. The formula is:

- Also the sulfur dioxide reacting with sodium hydroxide produces sodium sulfite and the formula is:

- And finally we have the reduction of sulfur dioxide (it can be in presence of hydrogen sulfide) getting elemental sulfur and water. The formula is:

Answer:
The answer to your question is below
Explanation:
For the reaction N2 + 3H2 = 2NH3, if you have 2.00 moles of N2 and 4.00 moles of H2 : a. Identify the limiting reactant. ____________
N2 + 3H2 = 2NH3
1 mol 3 moles 2 moles
Given 2 moles 4 moles
Rule of three for N2
x = (2 x 3) / 1 = 6 mol of H2, it means that is needed 6 moles of H2 and there are only 4, so the limiting reactant is H2
b. How many moles of NH3 can be formed? ____________
rule of three
3 moles of H2 --------------------- 2 moles of NH3
4 moles of H2 --------------------- x
x = (4x2)/3 = 2.7 moles of NH3 are formed
c. How many moles of the excess reactant will remain? _________
x = (4 x 1) / 3 = 1.33
Excess reactant = 2 - 1.33 = 0.67 moles
Heat energy is supplied to materials and can cause an increase in temperature of the material. the formula is as follows
H = mcΔt
where H - heat energy
m - mass of material
c - specific heat
Δt - change in temperature - 40.0 °C - 22.5 °C = 17.5 °C
substituting the values
H = 175 g x 0.234 Jg⁻¹°C⁻¹ x 17.5 °C
H = 716.6 J
716.6 J is required