Answer:
25 mL,50 mL, 20 mL
Explanation:
Molarity = numbers of mole / volume in liters
Glucose
stock solutions concentration = 1M
concentration needed = 0.05 M
volume of the preparation = 500ml = 500 / 1000 = 0.5l
number of moles of glucose in the solution = Molarity × molar mass = 0.05 × 0.5 L = 0.025 moles
volume of the stock needed = number of moles / molarity of the stock solution = 0.025 moles / 1 M = 0.025L = 25 mL
Asparagine
Molarity of the stock solution = 100mM = 0.1 M
Molarity needed = 10mM = 0.01 M
volume of the prepared solution = 0.5 L
number of moles in the prepared solution = 0.01 M × 0.5 L = 0.005 moles
volume of the stock solution needed = number of moles of the prepared solution / Molarity of the stock = 0.005 / 0.1 = 0.05L = 50 mL
NaH₂PO₄
Molarity of the stock solution = 50 mM = 0.05 M
Molarity of the needed = 2 mM = 0.002 M
Volume of the prepared solution = 0.5 L
number of moles in the prepared solution = 0.002 M × 0.5 L = 0.002 × 0.5 L = 0.001 moles
volume of the stock needed = 0.001 moles / 0.05 M = 0.02 L = 20 mL
It would be called an alloy.
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
B all objects move around the sun
<span>Cesium has the lowest electronegativity value. </span>
