<span>Melting is an endothermic process (i.e. it absorbs heat), whereas freezing is an exothermic process (i.e. it releases heat, or can be thought of, albeit incorrectly from a thermodynamics standpoint, as "absorbing cold"). The standard enthalpy of fusion of water can be used for both scenarios, but standard enthalpy is in units of energy/mass, so 10 times as much energy will be absorbed in the former scenario (melting 10 kg of ice) than what will be absorbed in the latter scenario (freezing 1 kg of water). For both processes, assuming the water is pure and at standard atmospheric pressure, and the entire mass remains at thermal equilibrium, the temperature of both the solid and the liquid will remain at precisely 0 degrees Celsius (273 K) for the duration of the phase change.</span>
In this question, the <span>patient needs to be given exactly 500 ml of a 5.0%. The content of the glucose should be:
</span>weight= volume * density* concentration<span>
500ml * 1mg/ml *5%= 25mg.
The </span><span>stock solution is 35%, then the amount needed in ml would be:
weight= volume * density* concentration
25mg= volume * 1mg/ml *35%
volume= 25/35%= 500/7= 71.43ml</span>
Answer is: 48,25 torr.
Raoult's Law: p = x(solv) · p(solv)
p - <span>vapour pressure of a solution.
</span>x(solv) - <span>mole fraction of the solvent.
</span>p(solv) - <span>vapour pressure of the pure solvent.
</span>n(ethanol) = 950g ÷ 46,07g/mol = 20,62 mol.
x(solv) = moles of solvent ÷ total number of moles
x(solv) = 20,62 ÷ 21,77 = 0,965.
p = 0,965 ·50,0 torr = 48,25 torr.
Answer:

Explanation:
Assume you are using 1 L of water.
Then you are washing 4 L of salty oil.
1. Calculate the mass of the salty oil
Assume the oil has a density of 0.86 g/mL.

2. Calculate the mass of salt in the salty oil

3. Calculate the mass of salt in the spent water

4. Mass of salt remaining in washed oil
Mass = 172 g - 150 g = 22 g
5. Concentration of salt in washed oil

Answer: 32.94 g
Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.
for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm
V = 15.0 L
T = 350.8 + 273 = 623.8 K
For krypton, 
n = 0.146 moles
for chlorine, 
n = 0.439
From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.
Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.
So, the amount of product formed is calculated from moles of krypton.
Molar mass of krypton tetrachloride is 225.61 gram per mol.
There is 1:1 mol ratio between krypton and krypton tetrachloride.

= 32.94 g of 
So, 32.94 g of the product will form.