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kow [346]
3 years ago
6

What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

l to reach the 2nd end point
Chemistry
1 answer:
Mkey [24]3 years ago
3 0

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

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How many moles of O₂ are needed to react completely with 35.0 mol of FeCl₃? *
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So we need to find the pounds of coffee beans required to produce 250 cups

so if 50 cups need 1 pound of coffee beans

therefore 250 cups need - 1/50 x 250 = 5 pounds of coffee beans

since we have to find the amount of coffee beans needed in kg, we have to convert pounds to kg

1 pound = 0.45 kg

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The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.085 M, how long would it take u
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Explanation:

The time after which 99.9% reactions gets completed is 40 minutes

Explanation:

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t = age of sample

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b)  Time taken for 0.085 M to decrease to 0.055 M

t=\frac{2.303}{0.053}\log\frac{0.085}{0.055}

t=8.2min

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What is the characteristic of the polar water molecule?
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