Answer: NaCl and H2O
Explanation: Products are on the right side of the arrow :)
Answer:
37.14 %
Explanation:
Using the equation, mass, M = D1 * V1
= D2 * V2
Where,
D1 = density of the liquid Nitrogen
D2 = density of gaseous Nitrogen
V1 = volume of the liquid Nitrogen
V2 = volume of the gaseous Nitrogen
Calculating V2,
0.808 * 185 = 1.15 * V2
Volume of Nitrogen after expansion = 129.98 m3.
Volume = L * b * h
= 10 * 10 * 3.5
Volume of the room = 350 m3.
Fraction of air = volume of Nitrogen after expansion/volume of the room * 100
= 129.98/350 *100
= 37.14 %
Answer:
5SiO2 + 2CaC2 → 5Si + 2CaO + 4CO2
Explanation:
When you are balancing an equation, you use coefficients to change the number of atoms of each type that are on each side of the equation.
You cannot change subscripts, the small numbers to the right of the substances. You can only add coefficients, the large numbers to the left of substances.
A balanced equation has the same number and type of atoms on both sides of the equation (the reactants on the left, the products on the right).
To begin, this equation has:
1 Si
2 O
1 Ca
2 C
on the left and
1 Si
1 Ca
3 O
1 C
on the right. Those numbers don't match!
By adding coefficients we end up with 5 Si, 10 O, 2 Ca and 4 C on both sides.
Answer:
BaI₂.5H₂O
Explanation:
Given Data:
Mass of Hydrated BaI₂ = 10.222 g
Mass of dried BaI₂ = 9.520 g
Mass of Water removed = 10.222-9.520 = 0.702 g
M.Mass of BaI₂ = 391.136 g/mol
M.Mass of Water = 18.02 g/mol
Now,
Calculate moles of dried BaI₂ as,
Moles = Mass / M.Mass
Moles = 9.520 g / 391.136 g/mol
Moles = 0.02434 moles
Calculate moles of Water as,
Moles = Mass / M.Mass
Moles = 0.702 g / 18.02 g/mol
Moles = 0.0389 moles
Then,
Calculate Mole ratio of BaI₂ and water as,
= 0.02434 moles BaI₂ / 0.0389 moles Water
= 0.625
Now,
We will convert this mole ratio to a whole number by multiplying it with a nearest integer,
= 0.625 × 8
= 5
Hence, this means for every one mole of BaI there are 5 moles of Water.
Result:
BaI₂.5H₂O