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2 years ago

58.0 g of K2SO4 was dissolved in 500 g of water. What is the molality of this solution?

Chemistry

1 answer:

podryga [215]2 years ago

7 0

Answer: The molality of solution is 0.66 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n\times 1000}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in g

moles of $K_2SO_4$ = $\frac{\text {given mass}}{\text {Molar Mass}}=\frac{58.0g}{174g/mol}=0.33mol$

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

$Molality=\frac{0.33\times 1000}{500g}=0.66mole/kg$

Therefore, the molality of solution is 0.66 mole/kg

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A gas occupies 18.5L at stp. what volume will it occupy at 735 torr and 57°c?

olga nikolaevna [1]

The volume that will be occupied at 735 torr and 57 c is 23.12 L

calculation

At STP temperature=273 k and pressure=760 torr

by use of combined gas formula

that is P1V1/T1= P2V2/T2

where; P1 =760 torr

T1= 273 K

V1= 18.5 L

P2= 735 torr

T2= 57+273= 330 K

V2=?

by making V2 the formula of subject

V2= T2P1V1/P2T1

V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L

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3 years ago

Balance Na2O+H2O → NaOH

AleksAgata [21]

Answer:
Na2O+H2O=2NaOH
Step by step exp.
Given:
Equation Na2O+H2O=NaOH
To find: Balance the equation
Solution:
Taking LHS of the equation
LHS=Na2O+H2O
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To balance the equation we have equal number of atom so we multply 2 to the RHS=2NaOH
There fore the equation form is
Na2O+H2O=2NaOH

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2 years ago

Which statement is true about both a solid and a gas? A. They both fill a container. B. The both have defined shapes in any spac

Fiesta28 [93]

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2 years ago

Read 2 more answers

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

Lilit [14]

Answer:

pH = 1.32

Explanation:

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

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2 years ago

Calculate the [OH-] [H+]=0.000025

Harman [31]

Explanation:

There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.

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2 years ago