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Lerok [7]
3 years ago
10

A container of gas has a volume of 3.5 L and a pressure of 0.8 atm. Assuming the temperature remains constant, what volume of ga

s would result if the pressure was 0.5 atm?
A. 5.6 L
B. 2.2 L
C. 0.2 L
D. 1.8 L
Chemistry
2 answers:
strojnjashka [21]3 years ago
7 0

Answer:

The correct answer is option A.

Explanation:

Initial volume of the gas =V_1=3.5 L

Final volume of the gas = V_2=?

Initial pressure of the gas =P_1=0.8 atm

Final volume of the gas = P_2=0.5 atm

Using Boyle's law:

P_1V_1=P_2V_2

0.8 atm \times 3.5 L=0.5 atm\times V_2

V_2=5.6 L

Hence,the correct answer is option A.

Harrizon [31]3 years ago
5 0

Answer : The correct option is, (A) 5.6 L

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas = 0.8 atm

P_2 = final pressure of gas = 0.5 atm

V_1 = initial volume of gas = 3.5 L

V_2 = final volume of gas = ?

Now put all the given values in the above equation, we get:

0.8atm\times 3.5L=0.5atm\times V_2

V_2=5.6L

Therefore, the final volume of the gas is 5.6 L.

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A solution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.15 ∘c. what is the molal concentration of glucose in this solution
Reil [10]
<span>1.16 moles/liter The equation for freezing point depression in an ideal solution is ΔTF = KF * b * i where ΔTF = depression in freezing point, defined as TF (pure) â’ TF (solution). So in this case ΔTF = 2.15 KF = cryoscopic constant of the solvent (given as 1.86 âc/m) b = molality of solute i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1. Solving for b, we get ΔTF = KF * b * i ΔTF/KF = b * i ΔTF/(KF*i) = b And substuting known values. ΔTF/(KF*i) = b 2.15âc/(1.86âc/m * 1) = b 2.15/(1.86 1/m) = b 1.155913978 m = b So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
7 0
3 years ago
If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make
vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

              2H₂  + O₂  -----------> 2H₂O

Now look at the reaction for mole ratio

             2H₂  + O₂  -----------> 2H₂O

             2 mole                       2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂  = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

                  2H₂         +       O₂        ----------->    2H₂O

          2 mole (2 g/mol)                                 2 mole (18 g/mol)

                    4 g                                                     36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

                       4 g of H₂ ≅ 36 g of H₂O

                        30 g of H₂ ≅ X of H₂O

Do cross multiplication

                  X of H₂O =  30 g x 36 g / 4 g

                  X of H₂O =  270 g

Now convert grams of H₂O into moles

               No. of moles = mass in grams/molar mass

Put values in above formula

               No. of moles = 270 g / 18 (g/mol)

               No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

5 0
3 years ago
Why are the top of sea mounts flat?
Firdavs [7]

Erosion from the waves wash away the top layer making it flat on top.

3 0
3 years ago
I need help on this please
tino4ka555 [31]

Answer:

The answer is c

Explanation:c

3 0
3 years ago
Consider the titration of 10.00 mL of a monoprotic weak acid with 0.1234 M NaOH. If the equivalence point volume of NaOH was det
gogolik [260]

The initial concentration of the unknown acid is 0.1900 M.

Explanation:

Titration is a chemical method of analysis to know the concentration and volume of the unknown chemical or analyte.

The formula for the titration is:

Macid x Vacid = Mbase x V base

The volume must be in litres. The volume is given in ml it should be divided with 1000 to obtain values in litre.

Data given are:

volume of acid= 10 ml 0.01 L

Molarity of the acid = ?

volume of the NaOH or base = 15.4 ml or 0.0154 L (equivalence point of the base)

molarity of the base = 0.1234 M

Applying the formula and putting the values, we get

Macid x 0.01 = 0.1234 x 0.0154

Macid =  0.1900 M

The weak acid is having molarity of 0.1900 M against the strong base with molarity of 0.1234M.

4 0
3 years ago
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