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elena-s [515]
3 years ago
12

Where did water become gas?

Chemistry
1 answer:
AlekseyPX3 years ago
3 0

Explanation:

hhojpk vk hl gp vo gv chhodti lbp p vighnon hl bbl hp hl jl

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At 293 K, methanol has a vapor pressure of 97.7 Torr and ethanol has a vapor pressure of 44.6 Torr. What would be the vapor pres
Bad White [126]

Answer: The vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

p_1=x_1p_1^0 and p_2=x_2P_2^0

where, x = mole fraction in solution  

p^0 = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_1+p_2p_{total}=x_Ap_A^0+x_BP_B^0

moles of ethanol=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{80g}{46g/mol}=1.7moles

moles of methanol= \frac{\text{Given mass}}{\text {Molar mass}}=\frac{97g}{32g/mol}=3.0moles

Total moles = moles of ethanol + moles of methanol = 1.7 +3.0 = 4.7

x_{ethanol}=\frac{1.7}{4.7}=0.36,

x_{methanol}=1-x_{ethanol}=1-0.36=0.64

p_{ethanol}^0=44.6torr

p_{methanol}^0=97.7torr

p_{total}=0.36\times 44.6+0.64\times 97.7=78.3torr

Thus the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.

8 0
4 years ago
Which of these is an example of a physical change?
miv72 [106K]
Sugar dissolving in warm water because when the sugar dissolves you can get the sugar back by letting the water evaporate.,
7 0
3 years ago
Read 2 more answers
How long will it take a piece of strontium-90 (produced in a nuclear test blast) weighing exactly 1000.0 g to be
Nonamiya [84]

Answer:

Time taken 172.8 years

Explanation:

Given data:

Total amount of strontium-90 = 1000.0 g

Amount left = 15.625 g

Half life of strontium = 28.8 years

Time taken = ?

Solution:

Number of half lives:

At time zero = 1000.0 g

At first half life = 1000.0g/2 = 500 g

At second half life = 500 g/2 = 250 g

At third half life = 250 g/ 2 = 125 g

At 4th half life = 125 g/ 2= 62.5 g

At 5th half life = 62.5 g/2 = 31.25 g

At 6th half life = 31.25 g / 2 = 15.625 g

Time taken:

Time taken = number of half lives × half life

Time taken = 6×28.8 years

Time taken = 172.8 years

7 0
3 years ago
A solution in which the maximum amount of solute is dissolved is said to be
ivanzaharov [21]

Saturated.

An unsaturated solution is a solution where more solute can still be added.

5 0
3 years ago
A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or
Pavlova-9 [17]

Answer:

(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.

(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.

(c) The expected number of fatalities on a average chemical plant are one in 12500 years.

Explanation:

Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.

In 300 years we have 600,000 hours.

With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.

If we have 2000 work hours a year, it is expected 0.0033 deaths/year.

1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year

The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).

In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:

FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8}  hours=1.67*10^{2} =167

A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.

If we assume FAR=4, as in an average chemical plant, we expect

4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}

This is equivalent to say

\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}

The expected number of fatalities on a average chemical plant are one in 12500 years.

4 0
3 years ago
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