Answer:
i am so sorry. i do not have a answer but i am trying to find questioms i can answer
2H₂₍g₎ + O₂ ₍g₎→ 2H₂O
138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O
128 mol H₂O
Answer:
4.85 x 10^25
Explanation:
thats what i was told by a calculator
Answer:
The answer to your question is: letter c (96%)
Explanation:
Indium -113 (112-9040580 amu) ₁₁₃In
Indium-115 (114.9038780 amu) ₁₁₅In
Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In
Formula
Atomic mass = m₁(%₁) +m₂(%₂) / 100
%₁ = x I established this is an equation
%₂ = 100 - x
Substituting values
114.82 = 112.8040x + 114.9039(100-x) /100 and know we expand and simplify
114.82 = 112.8040x + 11490.39 - 114.9039x /100
11482 = 112.8040x -114.9039x +11490.39
11482 - 11490.39 = 112.8040x -114.9039x
-8.39 = -2.099x
x = 3.99
Then % of Indium-115 = 100 - 3.99 = 96
Answer:
0.24M
Explanation:
The equation for the reaction is given below:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the equation above, we obtained the following information:
nA (mole of acid) = 1
nB (mole of base) = 2
Data obtained from the question include:
Va (volume of the acid) = 12mL
Ca (concentration of the acid) =?
Vb (volume of the base) = 36mL
Cb (concentration of the base) = 0.16 M
The Ca (concentration of the acid) can be obtained as follow:
CaVa/CbVb = nA/nB
Ca x 12 / 0.16 x 36 = 1 /2
Cross multiply to express in linear form as shown below:
Ca x 12 x 2 = 0.16 x 36
Divide both side by 12 x 2
Ca = 0.16 x 36/ 12 x 2
Ca = 0.24M
Therefore, the concentration of the acid is 0.24M
