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stiks02 [169]
4 years ago
14

"A 10.0 mL sample of this solution was transferred to a 500.0-mL volumetric flask and diluted to the mark with water. Then 10.0

mL of the diluted solution was transferred to a 250.0-mL flask and diluted to the mark with water. What is the final concentration of the KNO3 solution?"
Chemistry
1 answer:
Slav-nsk [51]4 years ago
3 0

Answer:

If the original solution is "X" M in KNO₃ (moles/Lt), then final solution will be in concentration (2/250)×X M

Explanation:

Original solution of KNO₃ is X M, then, we take a sample of 10 ml and transferre it to 500 ml:

1000ml=X moles KNO₃

10ml=1/100 X moles KNO₃, that will be in 500ml

We then take 10 ml of this new solution and transferre it to a flask with a final volume of 250ml, so:

500ml=(1/100) X moles KNO₃

10ml=(1/5000) X moles KNO₃, these moles will now be in 250 ml

With this, we can calculate the final concentration (in M, or moles/Lt):

250ml=(X/5000) moles KNO₃

1000ml=2/250×X M KNO₃

So, once you know the original concentration of KNO₃ solution, you should just multiply this number by (2/250) to get final concentration of 250 ml solution obtained by multiple dilutions

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A solution is prepared by combining 5.00 mL of 4.8x10-4 M NaSCN solution, 2.00 mL of 0.21 M Fe(NO3)3 solution and 13.00 mL of 0.
Tcecarenko [31]

Answer:

The analytical concentrations of thiocyanate ions:

[SCN^-]=0.00012 mol/L

The analytical concentrations of ferric ions:

[Fe^{3+}]=0.063 mol/L

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

1) Moles of sodium thiocyanate  = n

Volume of sodium thiocyanate solution = 5.00 mL = 0.005 L

(1 mL = 0.001L)

Molarity of the sodium thiocyanate = 4.8\times 10^{-4} M

n=4.8\times 10^{-4} M\times 0.005 L=2.4\times 10^{-6}mol

1 mole of sodium thiocyanate has 1 mol of thiocyante ions.

So, moles of thioscyanate ions in 2.4\times 10^{-6}mol  of NaSCN.

=1\times 2.4\times 10^{-6}mol=2.4\times 10^{-6}mol

2) Moles of ferric nitrate = n'

Volume of ferric nitrate solution = 2.00 mL = 0.002 L

Molarity of the ferric nitrate = 0.21 M

n'=0.002 M\times 0.21 L=0.00042 mol

1 mole of ferric nitrate has 3 moles of ferric ions.

So number of moles of ferric ions in 0.00042 moles of ferric nitrate is :

3\times 0.00042 mol=0.00126 mol

Volume of nitric acid = 13.00 mL

Total volume by adding all three volumes of solutions = V

V = 5.00 mL + 2.00 mL + 13.00 mL = 20.00 mL = 0.020 L

The analytical concentrations of thiocyanate ions:

[SCN^-]=\frac{2.4\times 10^{-6}mol}{0.020 L}=0.00012 mol/L

The analytical concentrations of ferric ions:

[Fe^{3+}]=\frac{0.00126 mol}{0.020 L}=0.063 mol/L

3 0
3 years ago
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Fill the three 3d subshells with the specified number of electrons.
lana66690 [7]

1. 5 electrons.

3d^{5}

  • Therefore, the 3d subshells blanks will be like this:
  • ↑ ↑ ↑ ↑ ↑

2. 6 electrons.

3d^{6}

  • The 3d subshells blanks will be:
  • ↑↓ ↑ ↑ ↑ ↑

3. 7 electrons.

3d^{7}

  • The 3d subshells blanks will be:
  • ↑↓ ↑↓ ↑ ↑ ↑

Hope you could understand.

If you have any query, feel free to ask.

8 0
2 years ago
Assume that an exhaled breath of air consists of 74.8% N2, 15.3% O2, 3.7% CO2, and 6.2% water vapour.
Goryan [66]

Answer:

a) PN₂ = 0.733 atm

PO₂ = 0.150 atm

PCO₂ = 0.036 atm

Pwater = 0.061 atm

b) 6.44x10⁻⁴ mol

c) 0.02 g

Explanation:

a) By the Dalton's Law, in a gas mixture, the total pressure is the sum of the partial pressures, and the partial pressure is the molar fraction of the gas multiplied by the total pressure.

PN₂ = 0.748*0.980 =0.733 atm

PO₂ = 0.153*0.980 = 0.150 atm

PCO₂ = 0.037*0.980 = 0.036 atm

Pwater = 0.062*0.980 = 0.061 atm

b) The number of moles of CO₂ can be calculated by the ideal gas law:

PV = nRT, where P is the pressure, V is the volume (0.455 L), n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (37°C + 273 = 310 K).

0.036*0.455 = 0.082*310*n

25.42n = 0.01638

n = 6.44x10⁻⁴ mol

c) For the combustion reaction of glucose:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

So, the stoichiometry is:

1 mol of glucose ------- 6 moles of CO₂

x ------- 6.44x10⁻⁴ mol of CO₂

By a simple direct three rule:

6x = 6.44x10⁻⁴

x = 1.073x10⁻⁴mol of glucose

Glucose has a molar mass equal to 180 g/mol, and its mass is the molar mass multiplied by the number of moles:

m = 180x1.073x10⁻⁴

m = 0.02 g

8 0
3 years ago
Krypton gas is 21 times denser than helium gas at the same temperature and pressure. which gas is predicted to effuse faster
Inessa [10]
Helium gas would be the one to effuse faster as compared to krypton. This is because krypton is heavier than helium. Heavier molecules would require more energy to move so they tend to effuse slower than molecules that are lighter which will only require less energy.
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What's the net force acting on an 18-kg object if an acceleration of 3.0 m/s2 is produced? A. 12 N B. 6 N C. 54 N D. 36 N
ASHA 777 [7]
Im sure the answer is c

6 0
3 years ago
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