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SIZIF [17.4K]
2 years ago
6

The ksp of tin(ii) hydroxide, sn(oh)2, is 5.45 × 10-27. calculate the molar solubility of this compound.

Chemistry
1 answer:
Murljashka [212]2 years ago
5 0
Let x be mol/l of Sn(OH)2 that dissolve 
Therefore, this will give x mol/l Sn2+ and 2x mol/l OH-
Ksp= (Sn2+)(OH-)^2 
      = (x)(2x)^2
      = 4x^3
4x^3 = 5.45 × 10^-27
      x = 1.11 × 10^-9
x = molar solubility = 1.11×10^-9 M





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The density of water is 1.00g/ml at 4 degrees Celsius.How many water molecules are present in 2.56ml of water at this temperatur
vivado [14]

Answer:

8.55x10^22 molecules

Explanation:

From the question given, the following data were obtained:

Density = 1g/mL

Volume = 2.56mL

Mass =?

Density = Mass /volume

Mass = Density x volume

Mass = 1 x 2.56

Mass = 2.56g

Now let us convert this mass (i.e 2.56g) of water to mole

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O = 2.56g

Number of mole of H2O=? Number of mole = Mass /Molar Mass

Number of mole of H2O = 2.56/18

Number of mole of H2O = 0.142mol

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of H2O contains 6.02x10^23 molecules.

Now if 1mole of H2O contains 6.02x10^23 molecules, then 0.142mol of H2O will contain = 0.142 x 6.02x10^23 = 8.55x10^22 molecules

6 0
3 years ago
Scoring Scheme: 3-3-2-1 Part I. How many mL of 6 M NaOH stock solution should be used to prepare the 500 mL of 0.2 M NaOH? Pleas
Orlov [11]

Answer : The volume of 6M NaOH stock solution is, 16.7 mL

Explanation :

To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of NaOH stock solution.

M_2\text{ and }V_2 are the molarity and volume of NaOH.

We are given:

M_1=6M\\V_1=?\\M_2=0.2M\\V_2=500mL

Putting values in above equation, we get:

6M\times V_1=0.2M\times 500mL\\\\V_1=16.7mL

Thus, the volume of 6M NaOH stock solution is, 16.7 mL

4 0
3 years ago
What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of hea
Ksju [112]

Answer:

A. 0.00420 J/(gi°C)

Explanation:

i got it rigt

8 0
2 years ago
A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled
dalvyx [7]

Answer:

His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.

Explanation:

The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.

Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.

Suppose the mass of the solute is m.

Originally, the density is = $\frac{m}{50}$     \left(\frac{\text{mass}}{\text{volume}}\right)

Now after adding extra 10 mL , the density becomes $\frac{m}{60}$.

Therefore, $\frac{m}{50}>\frac{m}{60}$

So the density decreases when we add more solution.

4 0
3 years ago
35.0 grams of nitrogen gas reacts with 60.0 grams of hydrogen gas: N2 + 3H2--> 2NH3
Misha Larkins [42]

Explanation:

Moles of N2 = 35.0g / (28g/mol) = 1.25mol

Moles of H2 = 60.0g / (2g/mol) = 30.0mol

Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.

Moles of NH3 = 1.25mol * 2 = 2.50mol.

Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.

30.0mol - 1.25mol * 3 = 26.25mol.

Excess mass of H2

= 26.25mol * (2g/mol) = 52.5g.

6 0
3 years ago
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