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Margarita [4]
2 years ago
14

How is this effect of dissolvedsubstances important in nature?​

Chemistry
1 answer:
dimaraw [331]2 years ago
8 0

Explanation:

Water is called the "universal solvent" because it is capable of dissolving more substances than any other liquid. This is important to every living thing on earth. It means that wherever water goes, either through the air, the ground, or through our bodies, it takes along valuable chemicals, minerals, and nutrients.

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the elements carbon hydrogen and oxygen are all part of the same blank on the periodic table........ i i think GROUP diangol row
Mrrafil [7]
The elements Carbon, Hydrogen, and Oxygen are all part of Non-Metals.

Hope This Helps! :)
6 0
3 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
2 years ago
A 2.26 M solution of KOH is prepared. Calculate the moles and mass of solute present in a 15.2-mL sample of this solution. The m
kenny6666 [7]

Answer:

0.0344 moles and 1.93g.

Explanation:

Molarity is defined as the ratio between moles of a solute (In this case, KOH), and the volume. With molarity and volume we can solve the moles of solute. With moles of solute we can find mass of the solute as follows:

<em>Moles KOH:</em>

15.2mL = 0.0152L * (2.26mol / L) = 0.0344moles

<em>Mass KOH:</em>

0.0344 moles * (56.11g/mol) = 1.93g of KOH

7 0
3 years ago
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

6 0
3 years ago
During photosynthesis, how many moles of water (H₂O) are needed to produce 150 grams of glucose
Nuetrik [128]

During photosynthesis, 5 moles of water are needed to produce 150 grams of glucose. The correct option is D.

<h3>What is photosynthesis?</h3>

It is the process by which green plants and some other organisms use sunlight to synthesize nutrients from carbon dioxide and water.

  • Step 1: Write the balanced equation for photosynthesis.

6 CO₂ + 6 H₂O ⇒ C₆H₁₂O₆ + 6 O₂

  • Step 2: Convert 150 g of C₆H₁₂O₆ to moles.

The molar mass of C₆H₁₂O₆ is 180.16 g/mol.

150 g × 1 mol/180.16 g = 0.833 mol

  • Step 3: Calculate the moles of water required to form 0.833 moles of C₆H₁₂O₆

The molar ratio of H₂O to C₆H₁₂O₆ is 6:1.

0.833 mol C₆H₁₂O₆ × 6 mol H₂O/1 mol C₆H₁₂O₆ = 5.00 mol H₂O

During photosynthesis, 5 moles of water are needed to produce 150 grams of glucose. The correct option is D.

Learn more about photosynthesis here: brainly.com/question/3529377

#SPJ1

6 0
2 years ago
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