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3 years ago

Do the gas laws apply to liquids ?

Chemistry

2 answers:

Contact [7]3 years ago

6 0

Answer:

The Ideal Gas Law cannot be applied to liquids. The Ideal Gas Law is #PV = nRT#. That implies that #V# is a variable. But we know that a liquid has a constant volume, so the Ideal Gas Law cannot apply to a liquid.

Explanation:

this is my awnser soory if it was a multiple choice question plz mark brainliest

tangare [24]3 years ago

3 0

the ideal gas law cannot be applied to liquids

the ideal gas law is PV= nRT

that implies that V is a variable. But we know that a liquid has a constant volume, so the Ideal Gas Law cannot apply to a liquid

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2 years ago

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7) Cooking or baking food results in a _________________ change

Flauer [41]

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3 years ago

How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?

zavuch27 [327]

Answer:

We need 12.26 grams H2SO4

Explanation:

Step 1: Data given

Volume of a H2SO4 solution = 500 mL = 0.500 L

Concentration of the H2SO4 solution = 0.250 M

Molar mass of H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = concentration * volume

Moles H2SO4 = 0.250 M * 0.500 L

Moles H2SO4 = 0.125 moles

Step 3: Calculate mass of H2SO4

Mass of H2SO4 = moles * molar mass

Mass of H2SO4 = 0.125 moles * 98.08 g/mol

Mass of H2SO4 = 12.26 grams

We need 12.26 grams H2SO4

7 0

3 years ago

A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO

Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL$

Putting values in above equation, we get:

$2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL$

$M_2=0.336M$

Thus, the concentration of NaOH is, 0.336 M

3 0

3 years ago

Bas + PtF2 → BaF2 + Pts Need to balance it

Basile [38]

it is already balanced

REACTANTS

Barium sulfide (BaS) + platinum (Ii) fluoride

PRODUCT

Barium fluoride (BaF2) + Cooperite (PtS)

Hope this answer helps you dear! take care

5 0

2 years ago