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3 years ago

As water is cooled from 4°C to 0° C, its density

Chemistry

1 answer:

Juliette [100K]3 years ago

8 0

Answer:

Option B - Density increases

Explanation:

Density is defined as:

δ = mass / volume

When T°2 is > T°1, volume has been increased

As the volume increased, density has decreased.

↓δ = mass / volume ↑

In the situation above, temperature decreased from 4° to 0°.

The volume has been reduced, so the density has been increased.

↑ δ = mass / volume ↓

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4) The initial rate of the reaction between substances P and Q was measured in a series of

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Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

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$rate = k[P]^{2} [Q]$

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Explanation:

Given rate of the reaction is:

$rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }$

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

$k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4$

Second row:

2. Rate value:

$rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1$

3.Third row:

$[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}$

4. Fourth row:

$[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}$

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3 years ago