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DiKsa [7]
3 years ago
8

Will a material with a mass of 110 g and a volume of 312 cm3 float on or sink in water?

Chemistry
1 answer:
joja [24]3 years ago
7 0

Answer:

Material is less dense than water so it will float in water.

Explanation:

Given data:

Mass of material = 110 g

Volume of material = 312 cm³

Will it float or sink in water = ?

Solution:

First of all we will calculate the density of material.

d = m/v

d = 110 g/ 312 cm³

d =  0.35 g/cm³

The density of water is 1 g/cm³ it means material is less dense than water so it will float in water.

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If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
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Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

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