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elena-14-01-66 [18.8K]
3 years ago
13

Question 12 A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The c

oncentration of a solution equals the mass of what's dissolved divided by the total volume of the solution. Here's how the student prepared the solution: The label on the graduated cylinder says: empty weight: 8.2g She put some solid sodium thiosulfate into the graduated cylinder and weighed it. With the sodium thiosulfate added, the cylinder weighed 76.148g . She added water to the graduated cylinder and dissolved the sodium thiosulfate completely. Then she read the total volume of the solution from the markings on the graduated cylinder. The total volume of the solution was 172.7mL . What concentration should the student write down in her lab notebook? Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
o-na [289]3 years ago
3 0

Weight of sodium thiosulfate   = 76.148 - 8.2

= 67.948 g.

Concentration of the solution = 67.948 / 172.7

= 0.393 g / mL.  to the nearest thousandth . (answer).

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A. is the correct point.

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2 years ago
What is the approximate percent by mass of potassium in KHCO3?
lozanna [386]

Answer:

The mass percent of potassium is 39%

Option C is correct

Explanation:

Step 1: Data given

Atomic mass of K = 39.10 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molar mass of KHCO3

Molar mass KHCO3 = 39.10 + 12.01 + 1.01 + 3*16.0

Molar mass KHCO3 = 100.12 g/mol

Step 3: Calculate mass percent of potassium (K)

%K = (atomic mass of K / molar mass of KHCO3) * 100%

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The mass percent of potassium is 39%

Option C is correct

8 0
3 years ago
A solution of 1.50 g of solute dissolved in 25.0 mL of H₂O at 25°C has a boiling point of 100.45°C. (d) Find the van’t Hoff fact
givi [52]

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<h3>What is the value of van 't Hoff factor?</h3>

For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

Which has highest Van t Hoff factor?

The Van't Hoff factor will be highest for

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To learn more about van 't Hoff factor off factor here:

brainly.com/question/22047232

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4 0
1 year ago
What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH
AlladinOne [14]
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2:    M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO:   M(CO) = 12.0 +16.0 = 28.0 g/mol

</span>12.0 g  H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO

<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover. 

                                  CO       +         2H2   ------->      CH3OH  
                                 1 mol              2 mol
given                        2.66 mol          6 mol (excess)

How much
we need  CO?           3 mol              6 mol

We see that H2 will be leftover, because for 6 moles H2  we need 3 moles CO, but we have only 2.66 mol  CO.
So, CO will react completely, and we are going to use CO to find  the mass of CH3OH.

3)                              </span>CO       +         2H2   ------->      CH3OH  
                                 1 mol                                        1 mol
                                2.66 mol                                    2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 +  4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH =  85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
3 0
3 years ago
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