The number of H atoms in 3(NH₄)₂CrO₄ = 24
<h3>Further explanation </h3>
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
Subscripts in the chemical formula indicate the number of atoms
The compound of 3(NH₄)₂CrO₄ ( 3 molecules of (NH₄)₂CrO₄ ) :
Number of H :
The mass of an atom and the equivalent of to the number of protons and neutrons in the atom
Answer:
1528.3L
Explanation:
To solve this problem we should know this formula:
V₁ / T₁ = V₂ / T₂
We must convert the values of T° to Absolute T° (T° in K)
21°C + 273 = 294K
70°C + 273 = 343K
Now we can replace the data
1310L / 294K = V₂ / 343K
V₂ = (1310L / 294K) . 343K → 1528.3L
If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too
Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
