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sukhopar [10]
3 years ago
5

An aqueous NaClNaCl solution is made using 111 gg of NaClNaCl diluted to a total solution volume of 1.20 LL .Calculate the molal

ity of the solution.
Chemistry
1 answer:
Aliun [14]3 years ago
7 0

Answer:

The answer to your question is m = 1.6

Explanation:

Data

mass of NaCl = 111 g

volume of water = 1.2 L

Molality = ?

Formula

Molality = number of moles / volume (L)

Process

1.- Calculate the moles of NaCl

Molecular mass = 23 + 35.5 = 58.5 g

                          1 mol of ------------------- 58.5 g

                          x           -------------------- 111 g

                          x = (111 x 1) / 58.5

                          x = 1.897 moles of NaCl

2.- Calculate the mass of water

density of water = 997 kg/m³

mass = volume x density

mass = 0.0012 x 997

mass = 1.19 kg

3.- Calculate molality

m = 1.897/1.19

m = 1.6

   

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A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?
Viktor [21]

The answer for the following problem is mentioned below.

  • <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules.</em></u>

Explanation:

Given:

mass of calcium phosphate (Ca_{3}(PO_{4} )_{2} ) = 125.3 grams

We know;

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = 120 + 3(95)

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} )  = 120 +285 = 405 grams

<em>We also know;</em>

No of molecules at STP conditions(N_{A}) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

\frac{m}{M} =\frac{N} }{}N÷N_{A}

\frac{405}{125.3} =\frac{N}{6.023*10^23}

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules</em></u>

3 0
3 years ago
Solve<br><br> 10N x 1m = 10)
stellarik [79]

Answer:

=10 joules

when multiplying with newtons ( the unit of force)

by a perimeter, it will becone joules, but the specific amount depends on how many newtons and metres there are

5 0
3 years ago
C3H8 + 5 O2 → 3 CO2 + 4 H2O
Reika [66]

Answer:

Mass = 112 g

Explanation:

Given data:

Mass of CO₂ produced = 90.6 g

Mass of oxygen needed = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂       →      3CO₂+ 4H₂O

Number of moles of CO₂:

Number of moles = 90.6 g/ 44 g/mol

Number of moles = 2.1 mol

Now we will compare the moles of  CO₂ and oxygen:

                 CO₂           :           O₂

                    3             :            5

                    2.1           :        5/3×2.1 = 3.5

Mass of oxygen needed:

Mass = number of moles × molar mass

Mass = 3.5 mol × 32 g/mol

Mass = 112 g

7 0
3 years ago
Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of
Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

5 0
4 years ago
Suppose you need of Grade 70 tow chain, which has a diameter of and weighs , to tow a car. How would you calculate the mass of t
BabaBlast [244]

Answer: check explanation

Explanation:

In this question we are to find mass. In order to calculate the Mass, We need the values of two parameters, that is, the values given for the grade tow chain, and the value given for the mass per length.

Assuming the mass per length is 3 Kilogram per metre(kg/m) and the grade 70 tow chain length is 5 metre(m).

Therefore, the formula for calculating mass of the chain is given below;

Mass of the chain= mass per unit length(kg/m) × length ---------------------------------------------------------------------------------------------------------------------(1).

Mass of the chain= 3 kg/m × 5 m.

Mass of the chain= 15 kg.

7 0
4 years ago
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