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3 years ago

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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3 years ago

A sample of mass 6.814 grams is added to another sample weighing 0.08753 grams.

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Answer:

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Explanation:

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3 years ago

A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a

lianna [129]

Answer:

P = 0.6815 atm

Explanation:

Pressure = 754 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 754 / 760 atm = 0.9921 atm

Temperature = 294 K

Volume = 3.1 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9921 atm × 3.1 L = n × 0.0821 L.atm/K.mol × 294 K

⇒n of helium gas= 0.1274 moles

Surface are = 1257 cm²

For a sphere, Surface area = 4 × π × r² = 1257 cm²

r² = 1257 / 4 × π ≅ 100 cm²

r = 10 cm

The volume of the sphere is :

$V=\frac {4}{3}\times \pi\times r^3$

Where, V is the volume

r is the radius

$V=\frac {4}{3}\times \frac {22}{7}\times {10}^3$

V = 4190.4762 cm³

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So, V (max) = 4.19 L

T = 273 K

n = 0.1274 moles

Using ideal gas equation as:

PV=nRT

Applying the equation as:

P × 4.19 L = 0.1274 × 0.0821 L.atm/K.mol × 273 K

P = 0.6815 atm

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3 years ago