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Degger [83]
3 years ago
14

How many mL of a .132 M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt?

Chemistry
1 answer:
Sever21 [200]3 years ago
4 0

Answer:

465mL

Explanation:

Volume of a solution, V =Mass of substance, m/(Molarity of the solution of the substance, M × molar mass of the substance, M.m)

Given in the question,

M=.132M

M.m=23+35.5 = 58.5g/mol

m=3.59g

V= 3.59/(.132×58.5)

V = 0.465L

Volume in mL = volume in L × 1000

= 0.465 × 1000 = 465mL

Therefore, 465mL of a .132M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt

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If you are given an ideal gas with pressure (p)259,392.00 pa and temperature (T)=200°c of 1 mole Argon gas in a volume 8.8dm3,ca
GuDViN [60]

Answer: R=4.82436 \frac{Pa. m^{3}}{mol. K}

Explanation:

The Ideal Gas equation is:  

P.V=n.R.T  (1)

Where:  

P is the pressure of the gas  

n the number of moles of gas  

R=8.3144598 \frac{Pa. m^{3}}{mol. K} is the gas constant  

T is the absolute temperature of the gas in Kelvin.

V is the volume

It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that <u>an ideal gas is a single hypothetical gas</u>. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be 8.3144598 \frac{Pa. m^{3}}{mol. K}.

However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.

Having this clarified, let's isolate R from (1):

R=\frac{PV}{nT}  (2)

Where:

P=259392 Pa

n=1 mole

T=200\°C=473.15 K is the absolute temperature of the gas in Kelvin.

V=8.8 dm^{3}=0.0088 m^{3}

R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}  (3)

Finally:

R=4.82436 \frac{Pa. m^{3}}{mol. K}  

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3 years ago
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A sample of CO2 gas has a volume of 2.7 L at 78.5 kPa. At what pressure would this sample of gas have a volume of 4.0L? Temperat
lisabon 2012 [21]

Answer:

52.99 kPa

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