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andrew-mc [135]

2 years ago

7

The concentration of hydrogen peroxide in a hair bleach is 0.050 mol dm–3. Use your answer from (b) to calculate the dilution fa

ctor needed to make the commercial hydrogen peroxide solution suitable for use in this hair bleach. Show your working.

1 answer:

stellarik [79]2 years ago

8 0

Answer:

0.03

Explanation:0.050 -

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In a laboratory setting, concentrations for solutions are measured in molarity, which is the number of moles per liter (mol/L).

slamgirl [31]

Answer:

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the stability of atomic nuclei is related to the _____. ratio of protons to electrons ratio of neutrons to protons number of pro

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In a post office, a 3-m long ramp is used to move carts onto a dock that is higher than 1 m. Which describes how the IMA of this

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2 years ago

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of

swat32

Ideal solutions obey Raoult's law, which states that:

P_i = x_i*(P_pure)_i

where
P_i is the partial pressure of component i above a solution
x_i is the mole fraction of component i in the solution
(P_pure)_i is the vapor pressure of pure component i

In this case,

P_benzene = 0.59 * 745 torr = 439.6 torr
P_toluene = (1-0.59) * 290 torr = 118.9 torr

The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:

P_total = (439.6 + 118.9) torr = 558.5 torr

Assuming the gas phase also behaves ideally, the partial pressure of each gas in the vapor phase is proportional to its molar concentration, so the mole fraction of toluene in the vapor phase is:

118.9 torr/558.5 torr = 0.213

8 0

2 years ago

A sample of 0.6760 g of an unknown compound containing barium ions (ba2+) is dissolved in water and treated with an excess of na

notka56 [123]

Answer: 35.72 % of Barium ions will be present in the original unknown compound.

Explanation: The reaction of Barium ions and sodium sulfate is:

$Na_2SO_4(aq.)+Ba^{2+}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)$

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.

Molar mass of Barium sulfate = 233.38 g/mol

Molar mass of Barium ions = 137.327 g/mol

233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

0.4105 grams of barium sulfate will be produced by = $\frac{137.327g/mol}{233.38g/mol}\times 0.4105g$ of Barium ions

Mass of barium ions = 0.2415 grams

To calculate percentage by mass, we use the formula:

$\% mass=\frac{\text{Mass of solute (in grams)}}{\text{Total mass of the solution(in grams)}}\times 100$

Mass of the solution = 0.6760 grams

Putting the value in above equation, we get

$\% \text{ mass of }Ba^{2+}\text{ ions}=\frac{0.2415g}{0.6760g}\times 100$

% mass of Barium ions = 35.72%.

8 0

2 years ago