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Xelga [282]
2 years ago
14

Is soap solution a strong, weak or non electrolyte

Chemistry
1 answer:
kramer2 years ago
4 0

Answer:

Nonelectrolytes do not dissociate into ions in solution; nonelectrolyte solutions do not, therefore, conduct electricity. ... Strong and Weak Electrolytes.

Explanation:

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4) How many grams are there in 7.40 moles of AgNO3
Ludmilka [50]

Answer:

1260 grams

Explanation:

3 0
3 years ago
What type of bond is made between amino acids in the ribosome?
sleet_krkn [62]

Answer:

<em><u>Peptide bonds form between the amino group of the amino acid attached to the A-site tRNA and the carboxyl group of the amino acid attached to the P-site tRNA. The formation of each peptide bond is catalyzed by peptidyl transferase, an RNA-based enzyme that is integrated into the 50S ribosomal subunit.</u></em>

8 0
2 years ago
A. 6.07 x 10-15<br> How many significant figures are there in this equation
Harman [31]

Answer:

3

Explanation:

In scientific notation, the number of integers in the first part will always be the amount of significant figures.

For example,

1) 1.542505 x 10^17 =7 significant figures

1) 1.54 x 10^17 =3 significant figures

8 0
3 years ago
Most strings of outdoor lights are wired in a parallel circuit. Which of the following is the best reason why a string of outdoo
sergey [27]

Answer:

C.If one light in the string burns out in a parallel circuit, the rest of the lights will continue to shine.

5 0
3 years ago
In lab (write this down in your lab protocol), you will be given a stock solution that has a glucose concentration of 60 mg/dL.
Wittaler [7]

Answer:

1. The dilution factor for the serial dilution = 2

2. V2 = 1 mL

3. V1 = 0.5 mL

Explanation:

1. Dilution factor is the ratio of the initial concentration to the final concentration.

Dilution factor = initial concentration / final concentration

First dilution: initial concentration = 60 mg/dL

final concentration = 30 mg/dL

Dilution factor = 60 mg/dL / 30 mg/dL = 2

Second dilution: initial concentration = 30 mg/dL

final concentration = 15 mg/dL

Dilution factor = 30 mg/dL / 15 mg/dL = 2

Therefore, the dilution factor for the serial dilution = 2

2. From the dilution formula, C1V1 = C2V2; V2 = final volume to be prepared.

Since 1 mL of the various glucose solutions are to be prepared, the final concentration, V2 = 1 mL

3. From the dilution formula, C1V1 = C2V2; V1 = initial concentration of the solution to be prepared.

C1/C2 = V2/V1

Since the dilution factor, C1/C2 is 2, V2/V1 = 2

V1 = V2/2

V1 = 1 mL / 2

V1 = 0.5 mL

6 0
3 years ago
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