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White raven [17]
3 years ago
11

The partial pressure of F​2​ in a mixture of gases where the total pressure is 1.0 atm, is 0.33 atm. What is the partial pressur

e of the other gases in the container?
Chemistry
1 answer:
olga nikolaevna [1]3 years ago
8 0

Answer:

Partial Pressure Formula

The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: Ptotal=P1+P2+… +Pn. + P n . The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.

Explanation:

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Calculate the energy required to melt 21 g of ice at 0 oC.
vladimir1956 [14]

<u>We are given:</u>

Mass of ice = 21 grams

The ice is already at 0°c, the temperature at which it melts to form water

Molar heat of fusion of Ice = 6.02 kJ/mol

<u>Finding the energy required:</u>

<u>Number of moles of Ice: </u>

Molar mass of water = 18 g/mol

Number of moles = given mass/ molar mass

Number of moles = 21 / 18 = 7/6 moles

<u>Energy required to melt the given amount of ice:</u>

Energy = number of moles * molar heat of fusion

Energy = (7/6) * (6.02)

Energy = 7.02 kJ OR 7020 joules

7 0
3 years ago
If 0.35 moles of kcl was dissolved in enough water to make 0.2 L of solution, what is molarity?
Mars2501 [29]

Answer:

1.75M

Explanation:

molarity = number of moles of solute/ number of L of solution =

=0.35 mol/0.2L = 1.75 mol/L = 1.75 M

8 0
3 years ago
1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules
Ivanshal [37]

Answer:

1. 2.510kJ  

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

1cal=4.184J\\\\1kJ=1000J

Then, we perform the conversion as follows:

600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ

2. Here, we use the general heat equation:

Q=mC(T_2-T_1)

And we plug in the given mass, specific heat and initial and final temperature to obtain:

Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ

Regards!

7 0
3 years ago
"determine the mass of oxygen" in a 7.9 g sample of al2(so4)3.
jeyben [28]

Answer:

              4.43 g of Oxygen

Explanation:

As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;

                          2 Moles of Aluminium

                          3 Moles of Sulfur

                          12 Moles of Oxygen

Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,

          342.15 g ( 1 mole) of Al₂(SO₄)₃ contains  =  192 g (12 mole) of O

So,

                         7.9 g of Al₂(SO₄)₃ will contain  =  X g of O

Solving for X,

                       X  =  (7.9 g × 192 g) ÷ 342.15 g

                      X =  4.43 g of Oxygen

7 0
3 years ago
Describe the position of metal ,non metals and metollieds in perodic tables​
siniylev [52]
Non metals and metollids in periodic tables are the same how this helps ;)
4 0
3 years ago
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