All categories

3 years ago

How many electron dots should As have?

Chemistry

1 answer:

Effectus [21]3 years ago

8 0

Answer:

Explanation:

You might be interested in

What is the pH value of Potassium hydroxide

RSB [31]

Answer:

from 10-14

Explanation:

the pH of potassium hydroxide is extremely high and is a strong base although the exact value depends on the concentration of the base in water.

3 0

2 years ago

Read 2 more answers

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0

3 years ago

The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen

tino4ka555 [31]

Answer: The mass percentage of $Tl_2SO_4$ is 5.86%

Explanation:

To calculate the mass percentage of $Tl_2SO_4$ in the sample it is necessary to know the mass of the solute ( $Tl_2SO_4$ in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the $TlI$ precipitate. We can establish a relation between the mass of $TlI$ and $Tl_2SO_4$ using the stoichiometry of the compounds:

$moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.$

Since for every mole of Tl in $TlI$ there are two moles of Tl in $Tl_2SO_4$ , we have:

$moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol$

Using the molar mass of $Tl_2SO_4$ we have:

$mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g$

Finally, we can use the mass percentage formula:

$mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%$

6 0

3 years ago

A 1.0 g sample of hydrogen reacts completely with 19.0 g of fluorine to form a compound of hydrogen and fluorine. a. What is the

Rashid [163]

Explanation:tr

a) Molar mass of HF = 20 g/mol

Atomic mass of hydrogen = 1 g/mol

Atomic mass of fluorine = 19 g/mol

Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{Molar mass of compound }}\times 100$

Percentage of fluorine:

$\frac{1\times 19 g/mol}{20g/mol}\times 100=95\%$

Percentage of hydrogen:

$\frac{1\times 1g/mol}{20 g/mol}\times 100=5\%$

b) Mass of hydrogen in 50 grams of HF sample.

Moles of HF = $\frac{50 g}{20 g/mol}=2.5 mol$

1 mole of HF has 1 mole of hydrogen atom.

Then 2.5 moles of HF will have:

$1\times 2.5 mol=2.5 mol$ of hydrogen atom.

Mass of 2.5 moles of hydrogen atom:

1 g/mol × 2.5 mol = 2.5 g

2.5 grams of hydrogen would be present in a 50 g sample of this compound.

c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.

Then mass of hydrogen in 50 grams of HF compound we will have :

5% of 50 grams of HF = $\frac{5}{100}\times 50 g=2.5 g$

8 0

3 years ago

Two steps in a synthesis of the analgesic ibuprofen include a carbonyl condensation reaction, followed by an alkylation reaction

Tresset [83]

Ibuprofen is synthesized by reacting ethyl 2-(4-isobutylphenyl)acetate with base, the base abstracts the acidic proton and enolate is formed which on reaction with diethyl carbonate generates diethyl 2-(4-isobutylphenyl)malonate (A). diethyl 2-(4-isobutylphenyl)malonate on treatment with Base again looses the acidic proton and forms enolate. The enolate with treatment with Methyl Iodide yields diethyl 2-(4-isobutylphenyl)-2-methylmalonate (B). diethyl 2-(4-isobutylphenyl)-2-methylmalonate on hydrolysis give Ibuprofen.

4 0

3 years ago

Read 2 more answers