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3 years ago

A beaker with water and the surrounding air are all at 24°C. After ice cubes are placed in the water, heat is transferred from:

2 answers:

7 0

The answer is (3) the water to ice cubes.

As the ice cubes should be at a temperature of about 0 degree ( freezing point) , at the same time the temperature of water is 24 degree. Thus, heat is transferred from water to ice cubes.

xxTIMURxx [149]3 years ago

5 0

(3) the water to the ice cubes
This is so because the ice cubes are colder compared to their surrounding objects.
They absorb the heat from the water, cooling it, therefore vanishing(melting).

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A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

3 years ago

In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m

salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : $Q=I\times t$

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec (1 min = 60 sec)

Now put all the given values in this formula, we get

$Q=10A\times 600s=6000C$

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = $2\times (1.6\times 10^{-19})$

Number of atoms deposited = $\frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22}$ atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = $\frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113$ moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g

Mass of Copper Deposited = $63.5\times 0.03113 =1.98g$

Therefore, the mass of copper deposit is, 1.98 grams

5 0

3 years ago

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\Delta T_f=i\times K_f\times m$