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Svet_ta [14]
3 years ago
11

What is the percentage by mass of sulphur in Al2(SO4)3[A12SO4 =342g/mol, S = 32]​

Chemistry
1 answer:
valentinak56 [21]3 years ago
4 0

Answer: The percentage by mass of sulphur in Al_2(SO_4)_3 is 9.36%

Explanation:

Mass percent of an element is the ratio of mass of that element by the total mass expressed in terms of percentage.

{\text {Mass percentage}}=\frac{\text {mass of sulphur}}{\text {Total mass}}\times 100\%

Given: mass of sulphur = 32 g/mol

mass of Al_2(SO_4)_3 = 342 g/mol

Putting in the values we get:

{\text {Mass percentage}}=\frac{32g/mol}{342g/mol}\times 100\%=9.36\%

The percentage by mass of sulphur in Al_2(SO_4)_3 is 9.36%

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The density of a metal is 10.5 g/cm3. If the mass of themetal
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Answer:

11.7 mL

Explanation:

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,  

d = density ,

From the question ,

The density of the metal = 10.5 g/cm³

The mass of the metal = 5.25 g

Hence , the volume can be calculated from the above formula , i.e. ,

d = m / V  

V = m / d

V = 5.25 g / 10.5 g/cm³

V = 0.5 cm³

Since , The unit 1 mL = 1 cm³

V = 0.5 mL

The vessel has 11.2 mL of water ,

The new volume becomes ,

11.2 mL +  0.5 mL = 11.7 mL

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3 years ago
At high pressures how does the volume of a real gas compare with the volume of an ideal gas under the same conditions and why
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Answer: No, a<span>t high pressures, volume of a real gas does not  compare with the volume of an ideal gas under the same conditions.

Reason: 
For an ideal gas, there should not be any intermolecular forces of interaction. However, for real gases there are intermolecular forces of interaction like dipole-dipole and dipole-induced dipole. Further, at high pressures, molecules are close by. Hence, extend of these intermolecular forces is expected to be high. This results in decreases in volume of real gas. Thus, </span>volume of a real gas does not  compare with the volume of an ideal gas under the same conditions.

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You have a ballon filled with hydrogen gas which keeps it at a constant pressure, regardless the volume. The initial volume of t
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Answer:

619°C

Explanation:

Given data:

Initial volume of gas = 736 mL

Initial temperature = 15.0°C

Final volume of gas = 2.28 L

Final temperature = ?

Solution:

Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)

Initial temperature = 15.0°C (15+273 = 288 K)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 2.28 L × 288 K / 0.736 L

T₂ = 656.6 L.K / 0.736 L

T₂ = 892.2 K

K to °C:

892.2 - 273.15 = 619°C

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