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A student was asked to prepare 500.0 mL of 6.0 M NaOH. The student measured 120.0 g of NaOH and placed it in a 1000 mL beaker. T

Morgarella [4.7K]

Answer:

The concentration of the solution will be much lower than 6M

Explanation:

To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.

From

n= CV

n = number of moles m/M( m= mass of solid, M= molar mass of compound)

C= concentration of substance

V= volume of solution

m=120g

M= 40gmol-1

V=500ml

120/40= C×500/1000

C= 120/40× 1000/500

C=6M

This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.

This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.

7 0

3 years ago

How does climate determine the organisms that live in a biome

victus00 [196]

Climate determines a biome’s characteristics by being the deciding factor in what can live and grow there.

5 0

3 years ago

48g of magnesium and 32g of oxygen react to form 80g of magnesium oxide.

ZanzabumX [31]

Mass of Oxygen : 8 g

<h3>Further explanation</h3>

Given

48g of Magnesium

32g of Oxygen

Required

Mass of Oxygen

Solution

The ratio of Mg and O in MgO :

48 g : 32 g = 3 : 2

Total ratio = 3+2 = 5

So for 20 g MgO, mass of Oxygen :

=2/5 x 20 g

=8 g

Or

%mass O = 32/80 x 100%=40%

mass O in 20 g MgO =

40% x 20 g =8 g

3 0

3 years ago

a) A water sample (density=1.00g/mL, S=0.28g/kg) contains Ca(2+) at a concentration of 42 mg/kg. Calculate the molarity of the i

natali 33 [55]

Answer : The molarity of the ion is $1.05\times 10^{-3}M$

Explanation : Given,

Density of sample = 1.00 g/mL

Concentration of $Ca^{2+}$ = 42 mg/kg

First we have to calculate the volume of sample.

Let us assume that the mass of sample be 1 kg or 1000 g.

$Density=\frac{Mass}{Volume}$

$1.00g/mL=\frac{1000g}{Volume}$

$Volume=\frac{1000g}{1.00g/mL}$

$Volume=1000mL=1L$ (1 L = 1000 mL)

Now we have to calculate the moles of $Ca^{2+}$

As we are given that,

Mass of $Ca^{2+}$ in 1 kg of sample = 42 mg = 0.042 g

Molar mass of Ca = 40 g/mole

$\text{Moles of }Ca^{2+}=\frac{\text{Mass of }Ca^{2+}}{\text{Molar mass of }Ca^{2+}}=\frac{0.042g}{40g/mol}=1.05\times 10^{-3}mol$

Now we have to calculate the molarity of the ion.

$\text{Molarity}=\frac{\text{Moles of }Ca^{2+}}{\text{Volume of sample}}$

$\text{Molarity}=\frac{1.05\times 10^{-3}mol}{1L}=1.05\times 10^{-3}mol/L=1.05\times 10^{-3}M$

Therefore, the molarity of the ion is $1.05\times 10^{-3}M$

8 0

3 years ago

Topic 1: Significant Figures

Eddi Din [679]

These are 6 questions and 6 answers, each with its explanation.

Answer:

a. 0.7540: 4 significant figures.

b. 12500: 3 significant figures.

c. 10000.01: 7 significant figures.

d. 1200: 2 significant figures.

e. 1.04 * 103: 3 significant figures.

f. 0.0080050: 5 significant figures.

Explanation:

a. 0.7540

When there are not digits (non-zero digits) before the decimal point, the significant digits are all the decimal digits from the first non-zero decimal. So, in this case, they are all the shown digits: 7, 5, 4, and 0, i.e. 4 significant figures.

b) 12500

When there is not decimal point, the significant digits include all the digits before only zeros appear. So, they are 1, 2, and 5, i.e. 3 significant figures.

When it is meant that all the digits are significan, a perod must be added, in this way: 12500. The last point is part of the number, it does not add value, but indicates that all the digits are significant.

c) 10000.01

This is the only number from the list with both decimal and non-decimal digits.

When there are digits before and after the decimal point, the significant digits include all the digits before the decimal point and all the decimal digits. So they are 1, 0, 0, 0, 0, 0, and 1, i.e. 7 significant digits.

d) 1200

Again, when there is not decimal point, the significant digits include all the digits before only zeros appear. So, they are 1 and 2, i.e. 2 significant figures.

The two last zeros do not count as significant digits. If the number were written 1200., i.e. showing a decimal point but without digits after the decimal point, then all the four digits were singnificant.

e) 1.04 × 10⁴

This number is written in scientific notation.

The numbers before the power of 10 are called mantissa.

For a number written in scientific notation, the significant digits are the digits in the mantissa, i.e. 1.04. So, the number of significant digits is 3 (1, 0, and 4).

f) 0.0080050

For a number with only decimal digits (there are not digits with value before the decimal point), the zeros after the decimal point and before the first non-zero digit do not count as significant figures. So, the significant figures are: 8, 0, 0, 5, and 0, i.e. 5 significant figures.

8 0

3 years ago