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Tasya [4]
3 years ago
8

1. A gas sample at a pressure of 5.00 atm has a volume of 3.00 L. If the gas pressure is changed to 760 mm Hg, what volume will

the gas occupy?
Chemistry
1 answer:
tatyana61 [14]3 years ago
8 0

Answer:

V₂ =  15.00 atm

Explanation:

Given data:

Initial pressure = 5.00 atm

Initial volume = 3.00 L

Final pressure = 760 mmHg ( 760/760 = 1 atm)

Final volume = ?

Solution:

P₁V₁ = P₂V₂

V₂ = P₁V₁ /  P₂

V₂ =  5.00 atm × 3.00 L / 1 atm

V₂ =  15.00 atm

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Of the concentration units below, only ________ uses kg of solvent in its calculation.
oksano4ka [1.4K]

Answer:

E) molality

Explanation:

Molality -

Molarity of a substance , is the number of moles present in a Kg of solvent .

Hence , the formula for molality is given as follow -

m = n / s

m = molality

s = mass of solvent in Kg ,

n = moles of solute ,

Hence , from the given information of the question,

The concentration unit which have Kg of solvent , is molality.

6 0
3 years ago
The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo
Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is 9.57\times 10^{-10}

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)  

                            3s                  s

The expression of K_{sp} for above equation follows:

K_{sp}=(3s)^3\times s

We are given:  

s=2.44\times 10^{-3}M

Putting values in above expression, we get:

K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}

Hence, the solubility product of silver (I) phosphate is 9.57\times 10^{-10}

4 0
3 years ago
What is the chemical formula of calcium chloride?
Luba_88 [7]
<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ It would be CaCl2

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0
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