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Arte-miy333 [17]
3 years ago
15

How many miles of a gas at 100 c does it take to fill a 1.00 l flask to a pressure of 152kPa

Chemistry
1 answer:
Hatshy [7]3 years ago
5 0

The complete question is as follows: How many moles of a gas at 100 c does it take to fill a 1.00 l flask to a pressure of 152kPa

Answer: There are 0.0489 moles of a gas at 100^{o}C is required to fill a 1.00 l flask to a pressure of 152kPa.

Explanation:

Given: Volume = 1.00 L,    

Pressure = 152 kPa (1 kPa = 1000 Pa) = 152000 Pa

Convert Pa into atm as follows.

1 Pa = 9.86 \times 10^{-6} atm\\152000 Pa = 152000 \times \frac{9.86 \times 10^{-6}atm}{1 Pa}\\= 1.5 atm

Temperature = 100^{o}C = (100 + 273) K = 373 K

Using the ideal gas formula as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\1.5 atm \times 1.0 L = n \times 0.0821 L atm/mol K \times 373 K\\n = \frac{1.5 atm \times 1.0 L}{0.0821 L atm/mol K \times 373 K}\\n = 0.0489 mol

Thus, we can conclude that there are 0.0489 moles of a gas at 100^{o}C is required to fill a 1.00 l flask to a pressure of 152kPa.

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nikklg [1K]

Answer:

THE MOLARITY IS 2.22 MOL/DM3

Explanation:

The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water

Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol

Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;

37.5 g of Na2S = 217 g of water

( 37.5 * 1000 / 217 ) g = 1000 g of water

So, 172.81 g/dm3 of the solution

So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass

Molarity = 172.81 g/dm3 / 78 g/mol

Molarity = 2.22 mol/dm3

The molarity of the solution is 2.22 mol/dm3

4 0
3 years ago
The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g
ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ

3 0
3 years ago
What is the mass of 18.0 mL of honey if its' density is 1.42 g/mL?
BigorU [14]

mass (m) = ? , volume = 18.0 ml , density = 1.42 g/ml .

density =  \frac{mass}{volume}  \\  \\ d =  \frac{m}{v}  \\  \\ mass = density \times volume \\  \\ m = d \times v \\  \\ m = 18 \: ml \times  \: 1.42 \:  \frac{g}{ml}  \\  \\ m = 25.56 \: g

I hope I helped you^_^

8 0
2 years ago
How long does it take me to travel 500 m E at a velocity of 50 m/s E?
konstantin123 [22]

Answer:15 s E

Explanation:

7 0
3 years ago
8. When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. I
Julli [10]

Hey there!:

8) ΔTb = i*Kb*m  

 m is molality

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m will be same for both

is 1 for glucose since it is covalent compound

is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻

So,  ΔTb will be 4 times in aluminum nitrate case

So, boiling point will change by 4ºC


9) use Q = m*  L

L =  heat of vaporization so:

T1=T2=100ºC

5.40 * 1000 => 5400  cal/g

Q =   5400 / 540

Q = 10 grams


Hope that thlps!

5 0
3 years ago
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