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Dmitrij [34]
3 years ago
13

What is the percent by mass of water in Na S04.10H20?

Chemistry
1 answer:
dalvyx [7]3 years ago
3 0

Answer:

                    Percent by mass of water is 56%

Explanation:

                    First of all calculate the mass of hydrated compound as,

Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g

Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g

Mass of Oxygen = O × 14 = 16 × 14 = 224 g

Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g

Mass of Na₂S0₄.10H₂O = 322.24 g

Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...

Mass of water = 10 × 18.02

Mass of water = 180.2 g

Now, we will apply following formula to find percent of water in hydrated compound,

           %H₂O  =  Mass of H₂O / Mass of Hydrated Compound × 100

Putting values,

                                      %H₂O  = 180.2 g / 322.24 g × 100

                                           %H₂O =  55.92 % ≈ 56%

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If a student has 125 mL of a 4.00 M CuSO4 solution and needs a 1.50 M solution, what volume do they need to dilute it to?
lara31 [8.8K]

Answer:

333.3mL

Explanation:

Using the formula as follows:

C1V1 = C2V2

Where;

C1 = initial concentration (M)

C2 = final concentration (M)

V1 = initial volume (mL)

V2 = final volume (mL)

According to the information provided in this question,

C1 = 4.00M

C2 = 1.50M

V1 = 125mL

V2 = ?

Using C1V1 = C2V2

4 × 125 = 1.5 × V2

500 = 1.5V2

V2 = 500/1.5

V2 = 333.3mL

Therefore, the CuSO4 solution needs to be diluted to 333.3mL to make 1.50 M solution.

3 0
2 years ago
Which sphere would this be found in?
andre [41]
That is correct…….. i think
8 0
2 years ago
Read 2 more answers
1.) A solute is separated from its solvent my evaporation. What kind of process is this?
RideAnS [48]
1) C !! Physical Process

2) A

3) D !! Water is universal solvent !

if you need any explanation ; comment !!
8 0
3 years ago
What volume did a helium-filled balloon have at 22.5 c and 1.95 atm if it’s new volume was 56.4 mL at 3.69 atm and 11.9c
Veseljchak [2.6K]

This is an exercise in the general or combined gas law.

To start solving this exercise, we obtain the following data:

<h3>Data:</h3>
  • T₁ = 22.5 °C + 273 = 295.5 K
  • P₁ = 1.95 atm
  • V₁ = ¿?
  • P₂ = 3.69 atm
  • T₂ = 11.9 °C + 273 =  284.9 k
  • V₂= 56.4 ml

We use the following formula:

P₁V₁T₂ = P₂V₂T₁ ⇒ General formula

Where

  • P₁ = Initial pressure
  • V₁ = Initial volume
  • T₂ = Initial temperature
  • P₂ = Final pressure
  • V₂ = final volume
  • T₁ = Initial temperature

We clear the formula for the initial volume:

\boldsymbol{\sf{V_{1}=\dfrac{P_{2}V_{2}T_{1}}{P_{1}T_{2}}  } }

We substitute our data into the formula to solve:

\boldsymbol{\sf{V_{1}=\dfrac{(3.69 \not{atm})(56.4 \ ml)(295.5 \not{k})}{(1.95 \not{atm})(284.9\not{k})}  }}

\boldsymbol{\sf{V_{1}=\dfrac{61498.278}{555.555} \ lm }}

\boxed{\boldsymbol{\sf{V_{1}=110.697 \ lm }}}

The helium-filled balloon has a volume of <u>110.697 ml.</u>

6 0
1 year ago
Read 2 more answers
Which one of the following changes would cause the pressure of a gas to double assuming temperature was held constant?
AleksandrR [38]

Answer: Increasing the volume of the gas would be the correct answer. Charles law of gas states this: 1) If the volume of a container is increased, the temperature increases.  

2) If the volume of a container is decreased, the temperature decreases.

Hope this helps

Explanation:

8 0
3 years ago
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