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Nesterboy [21]
3 years ago
5

What is the percent by mass in a solution with 10.0 grams of solute and 25.0 grams of solvent?

Chemistry
1 answer:
rewona [7]3 years ago
4 0

Answer:

Percent by mass of solute is (10/35) x 100 =  28.57%

Percent by mass of solvent is (25/35) x 100 = 71.43%

Explanation:

10.0 grams of solute

25.0 grams of solvent

Total Mass = 35 grams

percent by mass = (Mass A / Total Mass) x 100

Percent by mass of solute is (10/35) x 100 =  28.57%

Percent by mass of solvent is (25/35) x 100 = 71.43%

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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
Mkey [24]

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.

3 0
3 years ago
it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to
schepotkina [342]

Answer:

The right solution is "8.5\times 10^{-19} \ joule".

Explanation:

As we know,

1 mole electron = 6.023\times 10^{23} \ no. \ of \ electrons

Total energy = 513 \ KJ

                     = 513\times 1000 \ joule

For single electron,

The amount of energy will be:

= \frac{513\times 1000}{(6.023\times 10^{23})}

= 8.5\times 10^{-19} \ joule

8 0
3 years ago
Is C2H6 empirical molecular or both
marshall27 [118]

Answer:

molecular

Explanation:

7 0
3 years ago
An antacid tablet with an active ingredient of CaCO3 was dissolved in 50.0 mL of 0.300 M HCl. When this solution was titrated wi
yaroslaw [1]

Answer:

Explanation:

mole of HCl remaining after reaction with CaCO₃

= .3 M of NaOH of 32.47 mL

= .3 x .03247 moles

= .009741 moles

Initial HCl taken = .3 x .005 moles = .0015 moles

Moles of HCl reacted with CaCO₃

= .009741 - .0015 = .008241 moles

CaCO₃     +    2HCl   =   CaCl₂  +  CO₂  +  H₂O .

1 mole        2 moles

2 moles of HCl reacts with 1 mole of  CaCO₃

.008241 moles of HCl reacts with .5 x .008241 moles of  CaCO₃

CaCO₃ reacted with HCl =  .5 x .008241 = .00412 moles

the mass (in grams) of calcium carbonate in the tablet

= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )

5 0
3 years ago
a 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of
maw [93]
Hello there.

<span>A 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of HCL? what is the ph after the addition of. 0050 mol of NaOH?

Part 2 answer: </span><span>pH = 4.67 </span>
4 0
3 years ago
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