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Contact [7]
3 years ago
14

The forensic technician at a crime scene has just prepared a luminol stock solution by adding 16.0 g of luminol to H2O creating

a solution with a total volume of 75.0 mL. What is the molarity of the stock solution of luminol?
Chemistry
1 answer:
san4es73 [151]3 years ago
4 0

Answer : The concentration of luminol in the solution is, 1.20 M.

Explanation : Given,

Mass of luminol = 16.0 g

Volume of solution = 75.0 mL

Molar mass of luminol = 177.16 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of luminol}\times 1000}{\text{Molar mass of luminol}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{16.0g\times 1000}{177.16g/mole\times 75.0mL}=1.20mole/L=1.20M

Therefore, the concentration of luminol in the solution is, 1.20 M.

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What volume of 0.20 M solution of HCI will be required to neutralize 10.0 mL of a 0.60 M
tatyana61 [14]

Answer:

30ml

Explanation:

Chcl.Vhvl=Ckoh.Vkoh

4 0
2 years ago
When of a certain molecular compound X are dissolved in of dibenzyl ether , the freezing point of the solution is measured to be
kondaur [170]

This question is incomplete, the complete question is;

When 4.28 g of a certain molecular compound X are dissolved in 60.0 g of dibenzyl ether [(C₆H₅CH₂)₂0] , the freezing point of the solution is measured to be -3.2°C . Calculate the molar mass of X.

If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to significant digit.

Answer: molar mass of solute (X) is 88.03 g/mol

Explanation:

Given that;

mass of solute = 4.28 g

mass of solvent = 60.0 g = 0.060 kg        (Dibenzyl ether)

depression constant kf = 6.17 °CKg/mol

Freezing Point of solvent T₀ = 1.80°C       (Dibenzyl ether)

freezing point of solution Tsol = -3.20°C

Now we know that

Depression in freezing point ΔTf = depression constant kf × molaity m

and (ΔTf = T₀-Tsol)

so T₀ - Tsol = kf × m

we substitute

1.80 - (-3.20) = 6.17  × m

5 = 6.17 × m

m = 5 / 6.17

m = 0.8103 kg/mol

so molaity m = 0.8103 kg/mol

we know that

Molaity of solute m = (mass of solute / M.wt of solute) × ( 1 / mass of solvent in Kg)

solve for molar mass of solute

molar mass of solute =  (mass of solute / molaity) × ( 1 / mass of solvent in Kg)

now we substitute

molar mass = (4.28g / 0.8103 kg/mol) × (1 / 0.060kg)

molar mass = ( 5.2839 × 16.66 ) g/mol

molar mass = 88.0297 g/mol ≈ 88.03 g/mol

Therefore molar mass of solute (X) is 88.03 g/mol

3 0
3 years ago
The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the
Sonbull [250]

Answer:

The rate of consumption of NH_{3} is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

where -\frac{\Delta [NH_{3}]}{\Delta t} represents rate of consumption of NH_{3}, -\frac{\Delta [O_{2}]}{\Delta t} represents rate of consumption of O_{2}, \frac{\Delta [N_{2}]}{\Delta t} represents rate of formation of N_{2} and \frac{\Delta [H_{2}O]}{\Delta t} represents rate of formation of H_{2}O.

Here rate of formation of H_{2}O is 3.0 mol/(L.s)

From the above equation we can write-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Here \frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))

So, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Hence, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)  

6 0
3 years ago
What is the nuclear binding energy of an atom that has a mass defect of 5.0446 x 10-29 kg?
lutik1710 [3]
I think it is 4.54 x 10-12
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3 years ago
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siniylev [52]

Answer:

B

Explanation:

6 0
3 years ago
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