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3 years ago

100 ml is drawn from 0.1 M solution of KCl and added to 900 ml of water. What is the

Chemistry

1 answer:

grandymaker [24]3 years ago

3 0

Answer:

The new concentration will be 0.01 M.

Explanation:

To determine the new concentration we use the following formula.

concentration (1) × volume (1) = concentration (2) × volume (2)

concentration (1) = 0.1 M

volume (1) = 100 mL

concentration (2) = unknown

volume (2) = 100 mL + 900 mL = 1000 mL

concentration (2) = [concentration (1) × volume (1)] / volume (2)

concentration (2) = (0.1 × 100) / 1000 = 0.01 M

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15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

Calculate the molality for each of the following solutions. Then, calculate the freezing-point depression ΔTF = iKFcm produced b

zlopas [31]

Answer:

a) Cm= 3.9 m ; ΔTf= 14.51 ºC

b) Cm= 0.21 m ; ΔTf= 0.79ºC

Explanation:

In order to solve the problems, we have to remember that the molality (m) of a solution is equal to moles of solute in 1 kg of solvent.

m= mol solute/kg solvent

a) In this case we have molarity, which is moles of solute in 1 liter of solution. We have to know how many kg of solvent (water) we have in 1 L of solution.

3.2 M NaCl= 3.2 mol NaCl/ 1 L solution

1 L solution= 1000 ml solution x 1.00 g/ml= 1000 g

A solution is composed by solute (NaCl) + solvent, so:

1000 g solution = g NaCl + g solvent

g NaCl= 3.2 mol NaCl x 58.44 g/mol= 187 g NaCl

g solvent= 1000 g - 187 g NaCl= 813 g= 0.813 kg

Cm= 3.2 g NaCl/0.813 kg solvent= 3.9 m

NaCl is an electrolyte and it dissociates in water in two ions: Na⁺ anc Cl⁻, si the van't Hoff factor (i) is 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC/m x 3.9 m= 14.51ºC

b) In this case we have 24 g of solute in 1.5 L of solvent. We have to convert the liters of solvent to kg, and to convert the mass of solute to mol by using the molecular weight of KCl (74.55 g/mol):

24 g KCl x 1 mol KCl/74.55 g= 0.32 mol

1.5 L solvent= 1500 g solvent x 1.00 g/ml= 1500 g = 1.5 kg

Cm= 0.32 g KCl/1.5 kg solvent= 0.21 m

KCl is an electrolyte and when it dissolves in water, it dissociates in 2 ions: K⁺ and Cl⁻. For this, van't Hoff factor (i) is equal to 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC x 0.21 m= 0.79ºC

7 0

3 years ago

Identify oxidation.

nevsk [136]

Oxidation is "Increase in oxidation number" as well as loss of electrons.

A rise in oxidation number results from the loss of negative electrons, whereas a reduction in oxidation number results from the gain of electrons. As a consequence, the oxidized element or ion experiences a rise in oxidation number.

As a result of losing electrons in the process, a reactant oxidizes. When a reactant obtains electrons during a reaction, reduction takes place. This frequently happens when acid and metals react.

Therefore, Oxidation is "Increase in oxidation number" as well as loss of electrons.

Hence, the correct answer will be option (e)

To know more about Oxidation

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3 0

1 year ago

Gabe follows the directions to make chocolate chip cookies for his favorite Chemistry teacher. The recipe says it will make 24 c

Gemiola [76]

Answer:

71%

Explanation:

Theoretical Yield = 24

Actual Yield = 17

(Actual Yield/Theoretical Yield)*100% = (17/24)*100% ≈ 71%

7 0

3 years ago

Which two types of cells are involved in fertilization?

astra-53 [7]

Which two types of cells are involved in fertilization?

$\sf\purple{D.\: egg\: cell;\: sperm \:cell}$ ✅

Gametes are an organism's reproductive cells. The female gametes are called ova or egg cells, and the male gametes are called sperm.
During fertilisation, the gametes (an egg and sperm) fuse to form a zygote.

$\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘$

4 0

3 years ago