You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.
<em>Step 1.</em> Calculate the volume of 90 % alcohol needed
You can use the dilution formula
<em>V</em>1×<em>C</em>1 = <em>V</em>2×<em>C</em>2
where
<em>V</em>1 and<em> V</em>2 are the volumes of the two solutions
<em>C</em>1 and <em>C</em>2 are the concentrations
You can solve the above formula to get
<em>V</em>2 = <em>V</em>1 × <em>C</em>1/<em>C</em>2
<em>V</em>1 = 270 mL; <em>C</em>1 = 65 %
V2 = ?; _____<em>C</em>2 = 90 %
∴<em>V</em>2 = 270 mL × (65 %/90 %) = 195 mL
You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA
<em>Step 2</em>. Calculate the amount of water to add.
Volume of water = 270 mL – 195 mL = 75 mL
I'm not entirely sure but i think it's Se2-
Answer:
Mass of Ag produced = 64.6 g
Note: the question is, how many grams of Ag is produced from 19.0 g of Cu and 125 g of AgNO3
Explanation:
Equation of the reaction:
Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2
From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.
Molar mass of the reactants and products are; Cu = 63.5 g/mol, Ag = 108 g/mol, AgNO3 = 170 g/mol, Cu(NO3)2 = 187.5 g/mol
To determine, the limiting reactant;
63.5 g of Cu reacts with 170 * 2 g of AgNO3,
19 g of Cu will react with (340 * 19)/63.5 g of AgNO3 =101.7 g of AgNO3.
Since there are 125 g of AgNO3 available for reaction, it is in excess and Cu is the limiting reactant.
63.5 g of Cu reacts to produce 108 * 2 g of Ag,
19 g of Cu will react to produce (216 * 19)/63.5 g of Ag = 64.6 g of Ag.
Therefore mass of Ag produced = 64.6g
Answer:
a) r = k × [A] × [B]²
b) 3
Explanation:
Let's consider the following generic reaction
A + B + C ⇒ Products
The generic rate law is:
r = k × [A]ᵃ × [B]ᵇ × [C]ⁿ
where
This reaction is first order in A, second order in B, and zero order in C. The rate law is:
r = k × [A]¹ × [B]² × [C]⁰
r = k × [A] × [B]²
The overall order of the reaction is the sum of the individual reaction orders.
1 + 2 + 0 = 3
