C = 12 g
O = 16 g
H = 1 g
<h3>Further explanation
</h3>
Conservation of mass stated that
<em>In a closed system, the masses before and after the reaction are the same
</em>
we can calculate the mass of each atom in the compound :
O in O₂ :
mass O₂ = 32
mass O = 32 : 2 = 16 g
H in H₂O
mass H₂O = 18
mass 2.H + mass O = 18
mass 2.H + 16 = 18
mass 2.H=2
mass H = 1 g
C in CH₄
mass CH₄ = 16
mass C + mass 4.H = 16
mass C + 4.1=16
mass C = 12 g
or we can use formula :
Mass of a single C :
X4O10
Let molar mass of X be y
molar mass = 4y + 10 x 16 = 4y+160
so, moles = 85.2 / (4y+160)
Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0
so, 48 = 16 x 10 x [85.2 / (4y+160) ]
Solve the equation to get y.
y = 31
<span>The balance format is
4NH3+ 5O2 -------> 4NO + 6H2O </span>
Answer
solubility product = 3.18x 10^-7
Explanation:
We were given the pressure in torr then we need to convert to atm for consistency, ten we have
21torr/760= 0.0276315789 atm
21 Torr = .0276315789 atm
P = i M S T
M = P / iRT
Where p is osmotic pressure
T= temperature= 25C+ 273= 298K
for XY vanthoff factor i = 2
S = 0.0821 L-atm / mol K
M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)
M = 0.000564698046 mol/liters
solubility= 0.000564698046 mol/liters
Ksp = [X+][Y-]
Ksp = X^2
Ksp = [Sr^+2] * [SO4^-2]
Ksp = X^2
Ksp = (0.000564698046)^2
Ksp = 3.18883883 × 10-7
Ksp = 3.18x 10^-7
solubility product = 3.18x 10^-7
Therefore, the solubility product of this salt at 25 ∘C∘C is 3.18x 10^-7
