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artcher [175]
3 years ago
12

Determine the percent water in MgSO4*7H20?

Chemistry
1 answer:
Semmy [17]3 years ago
5 0

Answer:

\% H_2O=51.2\%

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol

Thus, the percent water is:

\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\

So we plug in to obtain:

\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%

Best regards!

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Salt is often added to water to raise the boiling point to heat food more quickly. if you add 30.0g of salt to 3.75kg of water,
sammy [17]

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

  • Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

  • Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

  • ΔT is the boiling-point elevation
  • Kb is the ebullioscopic constant.
  • b is the molality.
  • i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

7 0
2 years ago
4. How many atom are in <br> C3H5(NO3)3
shepuryov [24]

Answer:

carbon = 3

hydrogen=5

nitrogen=3

Oxygen=9

8 0
2 years ago
When particles that attract each other are allowed to come to together, energy is ______
Rufina [12.5K]

Answer:

Released

Explanation:

When particles that attract each other come together, energy is usually released. The combination of the two particles is expected to result in a lower energy system. This lower energy system will be more stable than the different individual particles.

Hence, as this lower energy system is formed, the excess energy originally possessed by the particles is evolved hence energy is released when particles that attract each other are allowed to come together.

5 0
3 years ago
For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati
Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

-N_A A = \frac{dm}{dt}

-N_A A = \frac{dVp}{dt}

-N_A A = \frac{dV}{dt}p

-N_A A = \frac{dhA}{dt}

-N_A A = \frac{dh}{dt} \times Ap

From the last step, area cancels out and thus leaves :

-N_A  = \frac{dh}{dt} \times p

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9}  \times (3 \times 10^{26}-C_{A2})=0.000246$$=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

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3 years ago
Define physical and chemical properties, provide three examples of each, discuss their reversibility, and explain the fundamenta
MakcuM [25]

Answer:

Physical properties are defined as the properties which can be observed without changing its chemical composition. For example: color, volume, and molecular weight. Chemical properties are defined as the properties which can be observed only after changing chemical identity of the substance.

<h3>Mark me a brainlist</h3>

5 0
3 years ago
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