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A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r

Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

5 0

2 years ago

A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung

BlackZzzverrR [31]

Answer:

0.053moles

Explanation:

Hello,

To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,

V = kN, k = V / N

V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx

V1 = 1.7L

N1 = 0.070mol

V2 = 1.3L

N2 = ?

From the above equation,

V1 / N1 = V2 / N2

Make N2 the subject of formula

N2 = (N1 × V2) / V1

N2 = (0.07 × 1.3) / 1.7

N2 = 0.053mol

The number of moles of gas in his lungs when he exhale is 0.053 moles

7 0

3 years ago

Is anyone else's brainly not working or is mine the only one that is tweaking?

omeli [17]

Mine tweaking as well

5 0

3 years ago

Which is a practical application of boiling-point elevation?

pishuonlain [190]

Boiling-point is the point of a pure liquid matter starts to evaporate and change into gaseous phase. It is where the set of conditions such as the pressure and temperature enough to do so. Boiling-point elevation, on the other hand, is the phenomenon of which the boiling point of a pure liquid matter is elevated because of the dissolved substances. A great example would be the boiling point of a distilled water (pure water) which is lesser than the boiling point of a sea water because of the dissolved salts. A pure water boils at 100°C at atmospheric pressure while a salt water boils at higher temperature than 100°C at the same pressure. Thus, the answer is D.

7 0

3 years ago

Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans

loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

Answer: The pH of the solution is 10.4

Explanation:

We are given:

Molarity of sodium hypochlorite = 0.39 M

$pK_a$ of HClO = 7.50