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Alla [95]
3 years ago
10

A sample of argon gas occupies 105 mL at 0.871 atm. If

Chemistry
1 answer:
qwelly [4]3 years ago
6 0

Answer:

1. final pressure = 0.259atm

2. 196.84mmHg

Explanation:

Using Boyle's law of equation

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (mL)

V2 = final volume (mL)

According to the information given in this question:

V1 = 105mL

V2 = 352mL

P1 = 0.871atm

P2 = ?

Using P1V1 = P2V2

P2 = P1V1/V2

P2 = 0.871 × 105/352

P2 = 91.455/352

P2 = 0.2598

P2 = 0.259atm

To convert 0.259atm of the gas into mmHg, we multiply the value in atm by 760.

Hence, 0.259 × 760

= 196.84mmHg

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Answer:

1. starting and stabilizing the stir function

2. a medium heat

3.turn up the heat setting

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

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When heating a solution to boiling on a hot plate, start by starting and stabilizing the stir function. Then, turn the heat to a medium heat to start. If necessary, turn up the heat setting after waiting about ten minutes without seeing boiling.

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3 years ago
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

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V_s = volume of solution in L

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moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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