Answer:
The answer is
<h2>4.0 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
<h3>
</h3>
From the question
mass of block = 36 g
volume = 9 cm³
The density is
We have the final answer as
4.0 g/cm³
Hope this helps you
<h3>
Answer:</h3>
Initial temperature is 243.59°C
<h3>
Explanation:</h3>
The quantity of heat is calculated by multiplying the mass of a substance by its specific heat capacity and change in temperature.
That is; Q = m×c×ΔT
In this case;
Quantity of heat = 560 J
Mass of the Sample of Zinc = 10 g
Final temperature = 100°C
We are required to determine the initial temperature;
This can be done by replacing the known variables in the formula of finding quantity of heat,
Specific heat capacity, c, of Zinc = 0.39 J/g.°C
Therefore,
560 J = 10 g × 0.39 J/g°C × ΔT
ΔT = 560 J ÷ (3.9 J/°C)
= 143.59°C
But, since the sample of Zinc lost heat then the temperature change will have a negative value.
ΔT = -143.59°C
Then,
ΔT = T(final) - T(initial)
Therefore,
T(initial) = T(final) - ΔT
= 100°C - (-143.59°C)
= 243.59°C
Hence, the initial temperature of zinc sample is 243.59°C
<em>Answer:</em>
<h3><em>Answer:</em><em> </em><em>well</em><em> </em><em> </em></h3>
<em>b. a type of gas is evolved ( hydrogen gas</em><em> )</em>
<em> </em>
Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.
Explanation :
As we know that:
At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.
The relation between pressure and number of moles of gas will be:
where,
= initial pressure of gas = 24.5 atm
= final pressure of gas = 5.30 atm
= initial number of moles of gas = 1.40 moles
= final number of moles of gas = ?
Now put all the given values in the above expression, we get:
Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
