Answer:
adding 750 grams of water at 60° Celsius
.
Explanation:
- We can calculate the amount of heat lost from 750 grams of water at 80°C to be lowered by 10°C using the relation:
<em>Q = m.c.ΔT,</em>
Where, Q is the amount of heat lost by water (Q = ??? J).
m is the mass of water (m = 750.0 g).
c is the specific heat capacity of the water (c = 4.18 J/g.°C).
ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.
Now, we can calculate the Q that is gained by the different added amounts of water:
- <em>adding 750 grams of water at 50° Celsius
:</em>
ΔT = 70.0°C - 50.0°C = 20.0°C,
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.
- <em>adding 325 grams of water at 60° Celsius
:</em>
ΔT = 70.0°C - 60.0°C = 10.0°C,
∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.
- <em>adding 750 grams of water at 60° Celsius
:</em>
ΔT = 70.0°C - 60.0°C = 10.0°C,
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.
- <em>adding 1000 grams of water at 55° Celsius:</em>
ΔT = 70.0°C - 55.0°C = 15.0°C,
∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.
<em>adding 750 grams of water at 60° Celsius</em>
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:
360 g : 1 L = x g : 30 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000
x = 10.8 g
So, from 30 mL mixture, 10.8 g of NaCl could be extracted.
Let's calculate the same for 10 mL water instead of 30 mL.
360 g : 1 L = x g : 10 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3
Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water. </span>
Answer:
189.71 secs
Explanation:
We know that decomposition is a first order reaction;
So;
ln[A] = ln[A]o - kt
But;
[A]o = 1.00 M
[A] = 0.250 M
t =135 s
Hence;
ln[A] - ln[A]o = kt
k = ln[A] - ln[A]o/t
k = ln(1) - ln(0.250)/135
k =0 - (-1.386)/135
k = 1.386/135
k= 0.01
So time taken now will be;
ln[A] - ln[A]o = kt
t = ln[A] - ln[A]o/k
t = ln (3) - ln(0.450)/0.01
t = 1.0986 - (-0.7985)/0.01
t = 1.0986 + 0.7985/0.01
t = 189.71 secs
