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1 year ago

5

select the correct answer from each drop down menu how do you the freezing point and boiling point of a solution compare with th

ose of us pure solvent? The freezing point of the solution is ___ and the boiling point is _____ than that of its pure solvent.

1 answer:

miss Akunina [59]1 year ago

4 0

Explanation:

select the correct answer from each drp down

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2 years ago

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

3 0

2 years ago

What is the mass, in grams, of 0.125 L of CO2 at STP? Question 4 options: 2.80 g 181 g 0.246 g 4.11 g

Zolol [24]

1) Use the fact that 1 mol of gas at STP occupies 22.4 liter

=> 1 mol / 22.4 l = x / 0.125 l => x = 0.125 l * 1 mol / 22.4 l = 0.00558 mol

2) Now use the molar mass of the gas

molar mass of CO2 ≈ 44 g / mol

Formula: molar mass = mass in grams / number of moles =>

mass in grams = molar mass * number of moles = 44 g/mol * 0.00558 moles

mass = 0.246 g

Answer: 0.246 g

7 0

2 years ago