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Leno4ka [110]
3 years ago
12

Question 1

Chemistry
1 answer:
olchik [2.2K]3 years ago
4 0

Answer:

Explanation:

nm

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How many ml of 2.50 M NaOH solution are required to make a 525 mL of 0.150 M?
larisa [96]

Answer:

31.5 mL of a 2.50M NaOH solution

Explanation:

Molarity (M) is an unit of concentration defined as moles of solute (In this case, NaOH), per liter of solvent. That is:

Molarity = moles solute / Liter solvent

If you want to make 525mL (0.525L) of a 0.150M of NaOH, you need:

0.525L × (0.150mol / L) = <em>0.07875 moles of NaOH</em>

<em />

If you want to obtain these moles from a 2.50M NaOH solution:

0.07875mol NaOH × (1L / 2.50M) = 0.0315L = <em>31.5 mL of a 2.50M NaOH solution</em>

7 0
3 years ago
Read 2 more answers
Which of the following has a mass of 10.0 g?
Serjik [45]

Answer:

a.  0.119mol Kr

Explanation:

To solve this problem, we must understand that;

     Mass  = number of moles x molar mass

Molar mass of Kr  = 83.3g/mol

                        Ar = 40g/mol

                        He = 4g/mol

                        Ne = 20.18g/mol

a0.119 mol Kr               mass  = 0.119 x 83.3 = 9.9g

b 0.400 mol Ar             mass  = 0.4 x 40  = 16g

C 1.25 mol He               mass  = 1.25 x 4  = 5g

d 2.02 mol Ne              mass  = 2.02 x 20.18  = 40.8

Krypton is the answer

5 0
3 years ago
I need help with homework
Aliun [14]
Number 9 is A and 10 is A
5 0
3 years ago
A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added
Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced.  The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = \frac{1.085 X 100}{1.45} =74.8 %.

7 0
3 years ago
the density of gold is 19.3 g/cm3. a 3.4 mg piece of gold is hammered into a square that is 8.6 x 10-6 cm thick. what is the len
faltersainse [42]

Answer:

The length of a side of the square is 4.53 cm

Explanation:

Volume (V) = mass/density

mass = 3.4 mg = 3.4/1000 = 0.0034 g

density = 19.3 g/cm^3

V = 0.0034/19.3 = 1.76×10^-4 cm^3

Volume = area × thickness

Area = volume/thickness = 1.76×10^-4/8.6×10^-6 = 20.48 cm^2

Length of one side of the square = sqrt(area of square) = sqrt(20.48 cm^2) = 4.53 cm

3 0
3 years ago
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