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3 years ago

What volume of a 0.33-m c12 h22 o11 solution can be diluted to prepare 25 ml of a solution with a concentration of 0.025 m?

1 answer:

3 0

Answer : When we are diluting a solution of $C_{12} H_{22} O_{11}$ from 0.33 m to 0.025 m of 25 ml we will need 1.89 ml of 0.33 m solution for the dilution.

Explanation : Using the formula $m _{1}V_{1}=m_{2} V_{2}

$

we can calculate the $V_{1} = (25 X 0.025) / 0.33 = 1.89 mL$

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A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

_______________________________________________

AnnyKZ [126]

Answer:

Already answered brainly.com/question/845321563

Explanation:

8 0

3 years ago

For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of

viktelen [127]

Answer:

$\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%$

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

$n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\$

Now, we compute the atomic percentages as shown below:

$\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%$

Best regards!

4 0

3 years ago

How do submarine volcanoes affect autotrophic bacteria?

Anarel [89]

Answer:A large number of autotrophic bacteria—bacteria that produce their own food—live near hydrothermal vents and submarine volcanoes.

These bacteria are considered chemosynthetic, meaning they produce food from chemical reactions usually involving carbon dioxide, oxygen, or hydrogen

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3 years ago

And ironic bonds what happens to electrons? No

iren [92.7K]

Answer:

metals donate electrons to nonsmetals

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4 years ago