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Agata [3.3K]
3 years ago
14

You are given a sample of a Group II metal chloride, but the identity of the metal is unknown. Knowing that all Group II metal c

hloride compounds have the general formula, MCI2 and that the percentage of chlorine in the compound is 34.1%, what is the identity of the unknown substance?​
Chemistry
1 answer:
topjm [15]3 years ago
4 0

Answer:

The identity of unknown substance is Barium.

Explanation:

The given compound has a formula of MCl2. And the percentage of chlorine in this compound is said to be 34.1 %. But, we know that the mass of chlorine in this compound is, 35.5*2 = 71 g

Therefore, if we let x be the total mass of the compound, then:

34.1% of x = 71 g

(0.341)x = 71 g

x = 71 g/0.341

x = 208.2 g

Hence, the mass of other atom M, must be:

M = x - 71 g

M = 208.2 g - 71 g

M = 137.2 g

Now, we look into periodic table in group II. We find that the element is Barium with atomic mass of 137 g.

<u>The identity of unknown substance is Barium.</u>

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At a festival, spherical balloons with a radius of 170.cm are to be inflated with hot air and released. The air at the festival
MissTica

Answer:

The correct answer is 1.5 balloons.

Explanation:

Based on the given question, the radius of the spherical balloons is 170 cm. Therefore, the volume of each balloon will be,  

= 4/3 × π × (170)³ cm³

= 20.5176 × 10⁶ cm³  

= 20.5176 m³

The density of air at 100 degree C s 0.946 Kg m⁻³

The mass of air in each balloon can be calculated by using the formula,  

Mass = density × volume

Mass = 0.946 Kg m⁻³ × 20.5176 m³

Mass = 19.410 Kg

The heat energy, that is, required to bring the air from 25 degree C to 100 degree C will be,  

= 19.410 × 10³ g × (100 -25) degree C × 1.009 J/g degree C

= 14.68 × 10⁵ J

The concentration of butane given is 1.00 Kg or 1000 g

The molecular weight of butane is 58.12 g per mole

The moles or n can be calculated by using the formula,  

n = mass / mol.wt

n = 1000 g/58.12 g/mol = 17.20 mol

The formation of enthalpy of butane at 25 degree C is 125.7 × 10³ J/mol. The evolution of heat energy that take place at the time of burning 17.20 moles of butane is,  

= 125.7 × 10³ J/mol × 17.20 mol

= 2.16 × 10⁶ J

The number of balloons that can be inflated with hot air is,  

= 2.16 × 10⁶ J / 14.68 × 10⁵ J per each balloon

= 1.5 balloons

Hence, maximum of 1.5 balloons can be inflated.  

7 0
3 years ago
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