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3 years ago

How many liters of plasma are present in 8.5 pints?

Chemistry

1 answer:

stellarik [79]3 years ago

4 0

Answer:

there are 4.0205 Litres of plasma in 8.5 pints

Explanation:

Recall that:

1 quart qt = 2 pints pt and 1 quart = 946 mL

∴ two pints will contain 946 mL

and one pint will be = 946 mL/2

= 473 mL

Similarly, 8.5 pints will contain 473 mL × 8.5 = 4020.5 mL

We know that :

1000 mL = 1 Litres

Hence 4020.5 mL = (4020.5 /1000)L

= 4.0205 Litres

Therefore, there are 4.0205 Litres of plasma in 8.5 pints

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2 years ago

Zn-64 = 48.63%

sladkih [1.3K]

Answer:

A = 65.46 u

Explanation:

Given that,

The composition of zinc is as follows :

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Zn-66 = 27.90%

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2 years ago

Given that the density of air is 14.4, then the vapour of sulphur IV oxide is A. 16 B. 32 C. 64 D. 128

Arisa [49]

Answer:

Explanation:

The vapour density of a gas is the number of times a given volume of gas or vapour is as heavy as the same volume of hydrogen at a particular temperature and pressure.

Vapour density = 2 × relative molecular mass of the gas or vapour

Relative molecular mass of SO2 = 32 + 2(16) = 64

Hence;

Vapour density of SO2 = 64/2

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6 0

2 years ago

What is the vapor pressure of a liquid at 305.03 K if its ∆Hvap = 28.9 kJ/mol and its normal boiling point is 341.88 K?

ASHA 777 [7]

<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm

<u>Explanation:</u>

To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm

$P_2$ = pressure of the liquid = ?

$\Delta H_{vap}$ = Heat of vaporization = 28.9 kJ/mol = 28900 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = 341.88 K

$T_2$ = final temperature = 305.03 K

Putting values in above equation, we get:

$\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm$

Hence, the vapor pressure of the liquid is 0.293 atm

5 0

2 years ago