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3 years ago

3) Calculate the percent by mass of 3.55 g NaCl dissolved in 88 g water.

Chemistry

1 answer:

kotykmax [81]3 years ago

5 0

Answer:

The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%

Explanation:

When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.

A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:

$Percent by mass=\frac{mass of solute}{mass of solution} *100$

In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.

In this case:

mass of solute: 3.55 g
mass of solution: 3.55 g + 88 g= 91.55 g

Replacing:

$Percentbymass=\frac{3.55}{91.55}*100$

Percent by mass= 3.88%

<u><em>The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%</em></u>

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for the following reaction how many grams of h20 will be produced if you react 3.7 grams of B2H6? B2h6 +3O2-2hbo2+ 2 H20

Nana76 [90]

Answer:

4.8 g H₂O

Explanation:

To find the mass of water, you need to (1) convert grams B₂H₆ to moles B₂H₆ (via molar mass from periodic table), then (2) convert moles B₂H₆ to moles H₂O (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles H₂O to grams H₂O (via molar mass from periodic table).

It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs because the given value (3.7 grams) has 2 sig figs.

Molar Mass (B₂H₆): 2(10.811 g/mol) + 6(1.008 g/mol)

Molar Mass (B₂H₆): 27.67 g/mol

1 B₂H₆ + 3 O₂ ---> 2 HBO₂ + 2 H₂O
^ ^

Molar Mass (H₂O): 15.998 g/mol + 2(1.008 g/mol)

Molar Mass (H₂O): 18.014 g/mol

3.7 g B₂H₆ 1 mole 2 moles H₂O 18.014 g
---------------- x --------------- x ----------------------- x ----------------- = 4.8 g H₂O
27.67 g 1 mole B₂H₆ 1 mole

6 0

2 years ago

What causes salinization

Svet_ta [14]

Answer:

Salinization is the increase of salt concentration in soil and is, in most cases, caused by dissolved salts in the water supply. This supply of water can be caused by flooding of the land by seawater, seepage of seawater or brackish groundwater through the soil from below.

Explanation:

8 0

3 years ago

Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write a balanced chemical equation for the reaction. (

Nesterboy [21]

Balanced chemical equation for the reaction is:

2S $O_{2}$ (g) + $O_{2}$ (g)+ 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

Moles of $H_{2} S_{} O_{4}$ formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of S $O_{2}$ = 11.5 moles

moles of $H_{2} S_{} O_{4}$ = ?

Moles of $O_{2}$ needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2S $O_{2}$ (g) + $O_{2}$ (g) + 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of S $O_{2}$ , x moles of $H_{2} S_{} O_{4}$ is formed

$\frac{1}{2} =\frac{x}{11.5}$

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1 mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

7 0

3 years ago

Write a method that could be used to produce pure crystals of copper chloride from copper oxide and hydrochloric acid.

joja [24]

- Describe how you would make the salt from the reactants.

CuO + 2 HCl → CuCl₂ + H₂O

- Describe how you would purify the salt from the reaction mixture.

Filter the solution and let the product crystallize.

Explanation:

The reaction between copper oxide (CuO) and hydrochloric acid (HCl) will produce copper chloride (CuCl₂) and water:

CuO + 2 HCl → CuCl₂ + H₂O

-Describe how you would make the salt from the reactants

In a beaker which contain cooper oxide add the hydrochloric acid. To avoid working with concentrated hydrochloric acid, the acid may be diluted with water but make sure the add the stoechiometric amount (and a little bit of excess) in respect with the copper oxide.

-Describe how you would purify the salt from the reaction mixture.

Let the reaction proceed and then filter the solution to remove the unreacted cooper oxide.

Let the filtered solution, which contain copper chloride, water and unreacted hydrochloric acid, to stand undisturbed for several days. You may see the crystals growing from the solution. To speed up the process you may reduce the temperature.

Learn more about:

purifying compounds

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3 years ago

Can someone tell me how do I finish this problem?!

Natalija [7]

You can cross multiply, it will be easy

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3 years ago