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nasty-shy [4]
2 years ago
6

The density of a 50% solution of naoh is 1.525 g/ml. what volume of a solution that is 50% by weight naoh is required to make 0.

4 liter of 0.1m naoh solution?
Chemistry
1 answer:
jolli1 [7]2 years ago
3 0
We consider that the 50% given in this item is by volume. Molarity is the unit of concentration that takes into account the number of moles of the solute and the number of moles of the substance.

      number of moles of NaOH = (0.1 moles / L)(0.4 L)
                      n = 0.04 moles of NaOH

If we are to take the basis of 1 mL of 50% NaOH solution, 
  
                  (1 mL solution)(1.525 g/mL)(0.50) = 0.7625 g
The number of moles is,
                  0.7625 g NaOH x (1 mol / 40 g) = 0.01906 moles of NaOH

                volume of solution required = (0.04 moles of NaOH)(1 mL solution / 0.01906 moles of NaOH)
                 
                 volume of solution required = 2.09 mL

Answer: 2.09 mL
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viktelen [127]

Answer: K_c for this reaction at this temperature is 0.029

Explanation:

Moles of  HBr = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M

The given balanced equilibrium reaction is,

                            2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial conc.              0.500 M              0  M        0 M  

At eqm. conc.            (0.500-2x) M   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}

Equilibrium concentration of [Br_2] = x =  0.0955 M

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

K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}

K_c=0.029

Thus K_c for this reaction at this temperature is 0.029

7 0
2 years ago
How many molecules of CBr4 are in 250 grams of CBr4
Kazeer [188]

Answer:- 4.54*10^2^3 molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


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For lysine, PI is:

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pH = \frac{1}{2} (9,0+12,5) = 10,75

At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.

Thus, <em>pH 9,8 is likely to work best for this separation</em>

<em></em>

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Adding 334Joules of heat to one gram of ice at STP will cause ice to change to water at the same temperature.

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Learn more:

Heat of fusion brainly.com/question/4050938

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