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DedPeter [7]
3 years ago
13

A smoke sample contains dust particles and several different gases, which have different combinations of molecules. the dust par

ticles are suspended in the gases. which term or terms could be used to describe this sample of smoke?
Chemistry
1 answer:
Vsevolod [243]3 years ago
7 0
The sample of smoke described above can be described as a heterogeneous mixture. This type of mixture do not have uniform properties and composition. So, getting a certain small sample would not represent the whole mixture since it does not have uniform composition. 
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What is formed when water chemically combines with carbon dioxide
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When water chemically combines with carbon dioxide, a Carbonic acid is formed.

<u>Explanation</u>:

  • Carbon dioxide responds with water in a solution to form a weak acid, carbonic acid. Carbonic acid disassociates into hydrogen particles and bicarbonate particles. The hydrogen particles and water respond with the most basic minerals modifying the minerals.  
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4 0
3 years ago
How many grams of solid NaOH are required to prepare a 400ml of a 5N solution? show your work!
Nuetrik [128]

<u>Answer:</u> The mass of solid NaOH required is 80 g

<u>Explanation:</u>

Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:

\text{Equivalent weight}=\frac{\text{Molecular weight}}{n}

where,

n = acidity for bases = 1 (For NaOH)

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

\text{Equivalent weight}=\frac{40g/mol}{1eq/mol}=40g/eq

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

Mathematically,

\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}

Or,

\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}         ......(1)

We are given:

Given mass of NaOH = ?

Equivalent mass of NaOH = 40 g/eq

Volume of solution = 400 mL

Normality of solution = 5 eq/L

Putting values in equation 1, we get:

5eq/L=\frac{\text{Mass of NaOH}\times 1000}{40g/eq\times 400mL}\\\\\text{Mass of NaOH}=80g

Hence, the mass of solid NaOH required is 80 g

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4 years ago
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