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AysviL [449]
3 years ago
8

An experiment was conducted with a 1.57 gram starting mass of copper. How many grams of cooper were collected if the yield was 8

0.6%? Report your answer with three significant figures.
Chemistry
1 answer:
IRINA_888 [86]3 years ago
7 0

Answer:

1.26 g of Cu

Explanation:

Since we have to find the percentage of the given amount so...

Yield in grams = (Initial mass used / 100) x Percentage Yield

Yield in grams = (1.57 / 100) x 80.6 = 1.26 g of Cu

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A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc
Sunny_sXe [5.5K]

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

7 0
3 years ago
Express in scientific notation. Make sure your answer has the same number of significant figures as the starting value. (6.0 x 1
Greeley [361]
I believe the correct answer from the choices listed above is the second option. The expression (6.0 x 104) (3.1 x 10-1) is equal to 1.9 x 10^4. We only have two significant figures from the starting expression that is why we rounded of the product from 1.86 to 1.9.
3 0
3 years ago
What are the boiling points and freezing points (in oC) of a solution of 50.3 g of I2 in 350 g of chloroform? The kb = 3.63 oC/m
patriot [66]

Answer:

Boiling point: 63.3°C

Freezing point: -66.2°C.

Explanation:

The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:

<em>Boiling point increasing:</em>

ΔT = kb*m*i

<em>Freezing point depression:</em>

ΔT = kf*m*i

ΔT are the °C that change boiling or freezing point.

m is molality of the solution (moles / kg)

And i is Van't Hoff factor (1 for I₂ in chloroform)

Molality of 50.3g of I₂ in 350g of chloroform is:

50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:

0.198 moles / 0.350kg = 0.566m

Replacing:

<em>Boiling point:</em>

ΔT = kb*m*i

ΔT = 3.63°C/m*0.566m*1

ΔT = 2.1°C

As boiling point of pure substance is 61.2°C, boiling point of the solution is:

61.2°C + 2.1°C = 63.3°C

<em>Freezing point:</em>

ΔT = kf*m*i

ΔT = 4.70°C/m*0.566m*1

ΔT = 2.7°C

As freezing point is -63.5°C, the freezing point of the solution is:

-63.5°C - 2.7°C = -66.2°C

7 0
3 years ago
How is energy involved in what happens during a chemical reaction
Semmy [17]
When the reactants have higher chemical change. 
7 0
3 years ago
How does a change in mass affect the motion of an object
Korolek [52]
The Chang in mass can control the motion due to mass added or taken away
5 0
3 years ago
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