Answer:
CH₄(g) + 2O₂(g) ---> 1CO₂(g) + 2H₂O(g)
Explanation:
any combustion of a hydrocarbon equation is in form:
CₓHₐ(g) + BO₂(g) ---> YCO₂(g) + ZH₂O(g), where x,a,b,y,z are all whole number positive integers
there will be 1 CO₂ to 2 H₂O, since there is 1 C to 4 H in CH₄; it is not 1:4 since 2 H is needed in H₂O
CH₄(g) + _O₂(g) ---> 1CO₂ + 2H₂O
there is 4 total O on products side, which can make 2O₂
CH₄(g) + 2O₂(g) ---> 1CO₂(g) + 2H₂O(g)
Answer:
molar mass CO = 28.01 g/mol
3.45 x 10^2 x 28.01 =9.66 x 10^3 g
mass Zn = 2.45 x 10^-2 x 65.39 g/mol=1.60 g => 1.60 x 10^0
Explanation:
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Answer:
2.5 mL of 1.0 x 10⁻⁴ M solution needs to be added
Explanation:
M₁ (Molarity after dilution) = 1.0 x 10⁻⁵ M
V₁ (volume after dilution) = 25.0 mL
M₂ (molarity before dilution) = 1.0 x 10⁻⁴ M
dilution equation is written as M₁V₁ = M₂V₂
Making V₂ the subject of formula
V₂ = M₁V₁ / M₂
Substitute the parameters in the equation,
V₂ = 1.0 x 10⁻⁵ x 25.0 / 1.0 x 10⁻⁴
= 2.5 mL
Volume of the solution before dilution is 2.5 mL
Repeat same step for the next
Gas water and ice i think bro