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3 years ago

What are the coefficients for the reaction _Cl2O5 + _H2O → _HClO3 once it is balanced?

Chemistry

1 answer:

Artemon [7]3 years ago

5 0

Cl2O5+H20-->2HClO3

Just remember to have the same number of each element on both sides of the equation.

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As the hydrogen gas was burning, a line spectra was observed, which is show below

Maksim231197 [3]

I'm sorry I think you forgot to post an image

3 0

3 years ago

A solution is prepared by adding 91.3 g of sodium bromide [NaBr], to 115. g of water. Determine whether the solution is saturate

MrMuchimi

Answer: The given solution is unsaturated solution

Explanation:

We are given:

Solubility of NaBr = 9.19 m

Unsaturated solution is defined as the solution in which more solute particles can be dissolved in the solvent.

Saturated solution is defined as the solution in which no more solute particles can be dissolved in the solvent.

A precipitate is defined as the insoluble salt which is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

There are three conditions:

When $m_{\text{(calculated)}}$ ; the solution is unsaturated
When $m_{\text{(calculated)}}=m_{\text{(given)}}$ ; the solution is saturated
When $m_{\text{(calculated)}}>m_{\text{(given)}}$ ; precipitate will form

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (sodium bromide) = 91.3 g

$M_{solute}$ = Molar mass of solute (sodium bromide) = 103 g/mol

$W_{solvent}$ = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

$\text{Calculated molality of NaBr}=\frac{91.3\times 1000}{103\times 115}\\\\\text{Calculated molality of NaBr}=7.71m$

As, the calculated solubility is less than the given solubility. So, the solution will be unsaturated.

Hence, the given solution is unsaturated solution

5 0

3 years ago

Prospectors are considering searching for gold on a plot of land that contains 2.45 g of gold per bucket of soil. If the volume

makvit [3.9K]

Answer:

13.2 g of gold

Explanation:

We'll begin by converting 5.25 L to ft³.

This can be obtained as follow:

Recall:

1 L = 0.0353 ft³

Therefore,

5.25 L = 5.25 × 0.0353

5.25 L = 1.85×10¯¹ ft³

From the question given above,

2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.

Therefore, Xg of gold will be present in 1 ft³ of soil i.e

Xg of gold = 2.45/1.85×10¯¹

Xg of gold = 13.2 g

Therefore, 13.2 g of gold is present in 1 ft³ of the soil.

8 0

4 years ago

Which of the following chemical formulas contains the most molecules?

maw [93]

C. 2Fe2O3 would be the correct answer!

4 0

3 years ago

According to Newton's Second Law of Motion, a decrease in mass will increase acceleration

elena55 [62]

is that the answer?????

8 0

3 years ago

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