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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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I believe the answer is A.
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Mr. Chem S. Tree added water to 250. ML of a 2.50 M NaOH solution, until the final volume was 500. ML. What is the new molarity
tresset_1 [31]

Answer:

molarity of diluted solution = 1.25 M

Explanation:

Using,          

C1V1 (Stock solution) = C2V2 (dilute solution)

given that

C1 = 2.50M

V1 = 250ML

C2 = ?

V2 = 500ML

2.50 M x 250 mL = C2 x 500 mL

C2 = (2.50 M x 250 mL) / 500 mL

C2 = 1.25 M

Hence, molarity of diluted solution = 1.25 M

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How many particles are in 2.0 moles of CH4
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Which compound is a hydrocarbon?
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O tetraclorometano é um solvente muito utilizado na indústria de transformaç?o de produtos orgânicos ele é produzido pela reaç?o
Vlada [557]

Answer:

ΔHr = -103,4 kcal/mol

Explanation:

<u>Using:</u>

<u>AH° (kcal/mol) </u>

<u>Metano (CH) </u>

<u>-17,9 </u>

<u>Cloro (CI) </u>

<u>tetraclorometano (CCI) </u>

<u>- 33,3 </u>

<u>Acido cloridrico (HCI) </u>

<u>-22</u>

It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:

ΔHr = (ΔH products - ΔH reactants)

For the reaction:

CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)

The balanced reaction is:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)

The ΔH's of formation for these compounds are:

ΔH CH₄(g): -17,9 kcal/mol

ΔH Cl₂(g): 0 kcal/mol

ΔH CCl₄(g): -33,3 kcal/mol

ΔH HCl(g): -22 kcal/mol

The ΔHr is:

-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)

<em>ΔHr = -103,4 kcal/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
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