Answer:
(b) Calculate the molarity of a solution of 4.8 mole of HCl in 600 mL of solution. ... (g) Calculate the mass of Na2CO3 that must be used to make 700 mL of a 0.136 M Na2CO3 ... (h) What mass of NaOH is needed to make 200 mL of a 0.300 M NaOH solution? ... However, when we are reacting solutions we have to convert.
Explanation:
Answer:
.125 M
Explanation:
.15 M/L * .125 L = .01875 moles
now dilute to 150 cc (by adding 25 cc)
.01875M / (150/1000) = .125M
When light strikes a transparent material, most of the light passes through it it doesn't absorb or reflect it
We can write the balanced equation for the synthesis reaction as
H2(g) + Cl2(g) → 2HCl(g)
We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) *
(2.02 g H2 / 1 mol H2)
= 4.056 g H2
We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
(70.91 g Cl2 / 1 mol Cl2)
= 142.4 g Cl2
Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
