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OverLord2011 [107]
2 years ago
10

8. What was the original concentration in the BHL sample, if the dilution is 1:500 and the concentration 0.07 mg/ml

Chemistry
1 answer:
Yakvenalex [24]2 years ago
4 0

Answer:

The original concentration is "35 mg/ml".

Explanation:

According to the question,

The solution is diluted,

= 1:50

The initial volume,

V1 = 1 ml

Final concentration,

= 0.07 mg

then,

The final volume,

V2 = 500 ml

As we know,

⇒ V_1N_1=V_2N_2

or,

⇒ N_1=\frac{V_2N_2}{V_1}

On substituting the values, we get

⇒       =\frac{500\times 0.07}{1}

⇒       =\frac{35}{1}

⇒       =35 \ mg/ml

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cluponka [151]

Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CuCl₂: 1 mole
  • NaNO₃: 2 moles
  • Cu(NO₃)₂ : 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

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  • NaNO₃: 85 g/mole
  • Cu(NO₃)₂ : 187.54 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
  • NaNO₃: 2 moles ×85 g/mole= 170 grams
  • Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
  • NaCl: 2 moles ×58.45 g/mole= 116.9 grams

<h3>Mass of NaCl formed</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?

mass of NaCl=\frac{2 grams of CuCl_{2}x116.9 grams of NaCl }{134.44grams of CuCl_{2}}

<u><em>mass of NaCl= 1.739 grams</em></u>

Finally, 1.729 grams of NaCl is formed.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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Rutherford theorized that atoms have their charge concentrated in a very small nucleus.

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That is different from Plum Pudding model, because it shows that most of the atom is empty space.

According to Rutherford model of the atom:

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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

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How many grams of lithium hypochlorite (LiClO) are there in 0.594 moles?
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Answer : The mass of lithium hypochlorite are, 34.7 grams.

Explanation : Given,

Moles of LiClO = 0.594 g

Molar mass of LiClO = 58.4 g/mol

Expression used :

\text{ Mass of }LiClO=\text{ Moles of }LiClO\times \text{ Molar mass of }LiClO

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

\text{ Mass of }LiClO=(0.594moles)\times (58.4g/mole)

\text{ Mass of }LiClO=34.7g

Therefore, the mass of lithium hypochlorite are, 34.7 grams.

3 0
3 years ago
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