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Schach [20]
3 years ago
10

Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap

Chemistry
1 answer:
Usimov [2.4K]3 years ago
5 0
Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ 
<span>1 mol of water weighs =  18.015g
</span>1.55 kg = <span>1550/18.015 mol = 86.03 mol 

</span><span>Heat required to vaporize :
</span>= 86.03 mol  x <span>40.7 kJ 
</span>
= 3501.421 kJ
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in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
Which statement best describes the main benefit to organisms that reproduce sexually?
8090 [49]

Answer:

B. They will conserve energy during reproduction

Explanation:

5 0
2 years ago
Read 2 more answers
A gas has a volume of 3L at 200 kPa. What will its volume be if the pressure is changed to 500 kPa?
marta [7]

Answer:

6L

Explanation:

<em>if it's 3L per 200kPa</em>

then it would be;

4L per 300kPa

5L per 400kPa

6L per 500kPa

that's how i'd work it out in my head, hope it helps, but not sure though!

5 0
3 years ago
. Which type of energy is involved in the operation of simple machines?
quester [9]
Input an output  energy
4 0
3 years ago
A colorless liquid has a molar mass of 60.01 g/mol. When the liquid was analyzed, it was 46.7% nitrogen and 53.3% oxygen. What i
MAXImum [283]

first we have to find the empirical formula of the compound

empirical formula is the simplest ratio of whole numbers of components making up a compound

for 100 g of the compound

N O

mass 46.7 g 53.3 g

number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol

moles = 3.34 mol = 3.33 mol

divide by the least number of moles

3.34/3.33 = 1.00 3.33/ 3,33 = 1.00

therefore number of atoms are

N - 1

O - 1

empirical formula is - NO


mass of empirical unit - 14 g/mol + 16 g/mol = 30 g

molecular formula is actual composition of elements in the compound

molecular mass - 60.01 g/mol


number of empirical units = molecular mass / empirical unit mass

= 60.01 g/mol / 30 g = 2

there are 2 empirical units


2(NO)

molecular formula = N₂O₂


6 0
3 years ago
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