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2 years ago

Calculate the amount of heat needed to evaporate 235.0 grams of water from 25.0°C to 100.0°C.

1 answer:

6 0

I hope this helps answer your question. :)

These problems are very hard to do digitally so I would recommend trying to practice these types of problems on paper :)

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5. A penny weighs about 2.5 g. How many moles of pennies would be required to equal the mass of the moon (7.3x10^24 kg)

schepotkina [342]

5. 1.16 x $10^{26}$ moles moles of pennies would be required to equal the mass of the moon.

6. 12.86 moles of ethanol are in a 750 ml bottle of vodka.

Explanation:

5 .Data given:

mass of penny = 2.5 grams

atomic mass of penny = 62.93 grams/mole

moles present in mass of the moon given as = 7.3 x $10^{24}$ kg

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles = $\frac{2.5}{62.93}$

0.039 moles of penny is present in 2.5 grams

0.039 moles of penny in 2.5 grams of it

so, x moles in 7.3 X $10^{27}$ grams

$\frac{0.039}{2.5} =\frac{x}{7.3 X 10^{27} }$

x = 1.16 x $10^{26}$ moles

so when the mass of the penny given is equal to the mass of moon, number of moles of penny present is 1.1 x $10^{26}$ .

Given:

vodka = 40% ethanol

volume of vodka bottle = 750 ml

moles of ethanol =?

density of ethanol =0.79 g/ml

atomic mass of ethanol = 46.07 grams/mole

so, from the density of ethanol given we can calculate how much ethanol is present in the solution.

density = $\frac{mass}{volume}$

density x volume = mass

0.79 x 750 = 592.5 grams

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles of ethanol = $\frac{592.5}{46.07}$

= 12.86 moles of ethanol

6 0

2 years ago

Can someone help me with this question ASAP

Mkey [24]

Aye i can’t see the full question dawg, you gotta send a full picture of it

6 0

1 year ago

PLZZ!!!- I NEED THIS RIGHT NOWA block is pulled 0.90 m to the right in 2.5 s.

UNO [17]

The average speed is 0.28

4 0

2 years ago

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

= 25kg

Mass ratio of oil to solution $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75= E₁ × N₁

75 = E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0

2 years ago

The smallest particle in any element is a compound. *

zubka84 [21]

Answer:

false.

Explanation:

the smallest particle of a element is an atom

7 0

2 years ago