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Vaselesa [24]
3 years ago
12

How many mL of a 4.50 (m/v)% solution would contain 23.0 g of solute?

Chemistry
2 answers:
nlexa [21]3 years ago
4 0
23.0 g/ (4.50 g/ mL) gives 5.11 mL as the volume.
Helen [10]3 years ago
3 0

<u>Answer:</u> The given amount of solute will be present in 511.11 mL of solution.

<u>Explanation:</u>

We are given:

Concentration of solution = 4.50 (m/v) %

This means that 4.50 grams of solute is present is 100 mL of solution.

We need to find the volume of solution that can contain 23.0 grams of solute.

By applying unitary method, we get:

4.50 grams of solute is present in 100 mL of solution.

So, 23.0 grams of solute will be present in = \frac{100mL}{4.50g}\times 23.0=511.11mL of solution.

Hence, the given amount of solute will be present in 511.11 mL of solution.

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The type of covalent bond is formed between amino acid molecules during protein synthesis will be  <u>"peptide bond".</u>

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2 years ago
25 POINTS! Will mark brainliest if you have a good response!
Zolol [24]
Electrons uniting with electrons of another atom is the cause in this relationship. The effect is a chemical change.
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What does the frequency of a wave represent? A. the distance between two consecutive crests B. the number of wave cycles that pa
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The frequency of a wave represents B. the number of wave cycles that pass through a specific point within a given time.

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Oxygen is a __________ and nitrogen is a __________. metalloid, metalloid nonmetal, metal nonmetal, nonmetal nonmetal, metalloid
Anna35 [415]

Answer:

"nonmetal, nonmetal"

Explanation:

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4 0
3 years ago
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
3 years ago
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