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2 years ago

Hi :) take some points for the road

Chemistry

2 answers:

Maurinko [17]2 years ago

6 0

Answer:

Explanation:

zhuklara [117]2 years ago

3 0

Answer:

oh rally? thank you >//<

Explanation:

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How are transuranium elements made?

g100num [7]

Explanation:

The transuranium elements are produced by the capture of neutrons

Hope it will help you

4 0

3 years ago

Read 2 more answers

Does anyone know how to do this ??

lord [1]

Answer:

Look at your periodic table

Explanation:

It tells you everything like Potasium 1 and 1 ox and 1 hydrogen

7 0

3 years ago

What fraction of a sample of ³H will be left after 36.78 years while undergoing beta decay with a half-life of 12.26 years?

frutty [35]

Explanation:

no lo c crack solo es para poder iniciar la sección y no c ingel Jaksjs

7 0

3 years ago

The mole fraction of a non-electrolyte (MM 40.0 g/mol) in a saturated aqueous solution is 0.310. What is the molality of the sol

jeka57 [31]

Answer: The molality of non-electrolyte is 24.69 m

Explanation:

We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

Where,

$n_{solute}$ = Moles of solute (non-electrolyte) = 0.310 moles

$W_{solvent}$ = Mass of solvent (water) = 12.42 g

Putting values in above equation, we get:

$\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m$

Hence, the molality of non-electrolyte is 24.69 m

4 0

2 years ago

An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Stels [109]

Answer: The percent yield of the reaction is 77.0 %

Explanation:

$2Pb+O_2\rightarrow 2PbO$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles$

$\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles$

According to stoichiometry:

2 moles of $Pb$ produces = 2 moles of $PbO_2$

2.18 moles of $Pb$ is produced by= $\frac{2}{2}\times 2.18=2.18moles$ of $PbO_2$

Mass of $PbO_2$ = $moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6$

percent yield = $\frac{374.7g}{486.6g}\times 100=77.0\%$

3 0

2 years ago