The answer
<span>the molar ratio for the following equation
____C3H8 + ____O2 Imported Asset ____CO2 + ____ H2O
</span>after it has been properly balanced:
__1_C3H8 + ____5O2 Imported Asset ____3CO2 + ____ 4H2O
proof:
number of C =3 (C3H8; 3CO2)
number of H =8 (C3H8 ; 4H2O)
number of O = 10(5x2) or (2x3+4) (5O2;4H2O)
the answer is
<span>Reactants: C3H8 = 1, O2 = 8; Products: CO2 = 3 and H2O = 4</span>
0.5
Explanation:
Given parameters:
Mass of Ca²⁺ = 10g
unknown:
Equivalent weight = ?
Solution:
Equivalent weight that is the amount of electrons which a substance gains or loses per mole.
Ca²⁺ has +3 charge
It lost 2e⁻;
therefore;
In 1 mole of Ca²⁺, we have 2 equivalent weight
1 mol Ca²⁺ = 2eq. wts.
1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt.
Therefore;
10.0 g Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.
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Molar mass brainly.com/question/2861244
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The substance is followed by H2O
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.
1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.
2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>
</span>