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Nikitich [7]
3 years ago
15

My theoretical yield of beryllium chloride was 10.7 grams. If my actual yield was 4.5 grams, what was my percent yield?

Chemistry
1 answer:
artcher [175]3 years ago
6 0

Answer:

The percent yield of this reaction is 42.1 % (option B)

Explanation:

Step 1: Data given

Theoretical yield of beryllium chloride = 10.7 grams

actual yield of beryllium chloride = 4.5 grams

Step 2: Calculate the percent yield for this reaction

Percent yield = (actual yield / theoretical yield) * 100%

Percent yield = (4.5 grams / 10.7 grams ) * 100 %

Percent yield = 42.1 %

The percent yield of this reaction is 42.1 % (option B)

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A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe
Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: V_{1} = 571 mL,       T_{1} = 26^{o}C

(a) T_{2} = 5^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL

(b) T_{2} = 95^{o}F

Convert degree Fahrenheit into degree Cesius as follows.

(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL

(c) T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL

8 0
3 years ago
A high concentration of hydrogen ions means a solution is what?
otez555 [7]

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6 0
3 years ago
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Qual é o principal fator que indica que o organismo está adptado?
Hoochie [10]

Answer:

Explanation:

Any genetically based physiological, behavioral, or ecological trait that enables an organism to cope with, and to survive and reproduce in, its environment represents an adaptation.

Qualquer traço fisiológico, comportamental ou ecológico de base genética que permite a um organismo enfrentar, sobreviver e se reproduzir em seu ambiente representa uma adaptação.

5 0
3 years ago
How many calories of heat were added to 347.9 g of water to raise its temperature from 25oC to 55oC?
saw5 [17]

Answer:

10437calories

Explanation:

The following data were obtained from the question given:

M = 347.9g

C = 4.2J/g°C

T1 = 25°C

T2 = 55°C

ΔT = 55 — 25 = 30°C

Q =?

Q = MCΔT

Q = 347.9 x 4.2 x 30

Q = 43835.4J

Converting this to calories, we obtained the following:

4.2J = 1 calorie

43835.4J = 43835.4/ 4.2 = 10437calories

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Question 2 of 25
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The right objective is A I guess

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