(a) Dilution formula, M1 V1 = M2 V2 2.90 * 6.00 = M2 * 175 M2 = final concentrat

Question

(a)

Dilution formula,

M1 V1 = M2 V2

2.90 * 6.00 = M2 * 175

M2 = final concentration = 0.0994 M

(b)

Moles of AgNO3 = molarity * volume / 1000 = 0.751 * 66.0 / 1000 = 0.0496 mol

From the balanced equation,

2 mol of AgNO3 can form 1mol of Ag2CO3

Then,

0.0496 mol of AgNO3 forms = 0.0496 * 1 / 2 = 0.0258 mol of Ag2CO3

Mass of Ag2CO3 = moles * molar mass = 0.0258 * 275.7 = 7.11 g.

(c)

moles of MgCO3 = mass / molar mass = 5.61 / 84.0 = 0.0668 mol

From the balanced equation,

1 mol of MgCO3 needs 2 mol of HNO3

then,

0.0668 mol of MgCO3 needs 2 * 0.0668 =0.134 mol

So, Volume = moles * 1000 / molarity

V = 0.134 * 1000 / 0.689

V = 194.5 mL

Explanation / Answer

Use the Refereaces to access Important valses if seeded for this question. In the laboratory you dilute 2.90 ml of a concentrated 6.00 M hydrobromic acid solution to a total volume of 175 mL. What is the concentration of the dilute solution? Submit Answer Retry Entire Group 9 more group attempts remaining Use the References to access important values if needed for this question How many grams of Ag COs will precipitate when excess K CO, solution is added to 66.0 mL of 0.751 M AgNOj solution? Submit Answer Retry Entire Group 9 more group attempts remaining Use the References to access important values if needed for this question. How many mL of 0.689 M HNOj are needed to dissolve 5.61 g of MgCO,? 2HNOj(aq)+ MgCO(s)Mg(NOj)z(aq)+ H20)+ CO.(8) mL Submit Answer Retry Entire Group 9 more group attempts remaining

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