From the calculations, the concentration of the acid is 0.24 M.
<h3>What is neutralization?</h3>
The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.
We have to use the formula;
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O
CA = ?
CB = 1.2 M
VA = 50 mL
VB = 20 mL
NA = 1
NB = 2
CA = CBVBNA/VANB
CA = 1.2 M * 20 mL * 1/ 50 mL * 2
CA = 0.24 M
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Answer:
34g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H2S + 2AgNO3 —> 2HNO3 + Ag2S
Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.
This is illustrated below:
From the balanced equation above,
We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.
Finally, we shall convert 1 mole of H2S to grams. This is shown below:
Number of mole H2S = 1 mole
Molar mass of H2S = (2x1) + 32 = 34g/mol
Mass = number of mole x molar Mass
Mass of H2S = 1 x 34
Mass of H2S = 34g
Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.
Given that the volume and amount of water are kept constant,
P/T = constant
P₁/T₁ = P₂/T₂
Normal atmospheric pressure is 746 mmHg and normal boiling point of water is 100 °C.
746/100 = 589/T₂
T₂ = 79.0 °C
Answer:
Na₂₆F₁₁
Explanation:
We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:
74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na
25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F
Divide each by smallest number of moles:
3.2255/1.3586 = 2.37
1.3586/1.3586 = 1
Multiply by common number to get a smallest whole number:
2.37*11 = 26,
1*11 = 11
The empirical formula is Na₂₆F₁₁
