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3 years ago

5

What effect would the incomplete drying of a sample (for example, the incomplete removal of a re-crystallization solvent) on per

cent recovery?

1 answer:

Varvara68 [4.7K]3 years ago

5 0

Answer:

It will lead to overestimation of the percent recovery

Explanation:

The percent recovery refers to the percentage of the pure compound recovered after a chemical process. It is a ratio of the pure compound recovered to the original substance multiplied by 100%.

If the sample was incompletely dried and the recrystallization solvent is still present, the percent recovery will be overestimated and we will have a value that is greater than the accurate percent recovery due to solvent impurities present.

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A 33.153 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap

elena-14-01-66 [18.8K]

Answer:

The empirical formula of the compound = $C_4H_8S_1O_1$

Explanation:

Mass of carbon dioxide gas = 59.060 mg = 0.059060 g

1 mg = 0.001 g

Moles of carbon dioxide = $\frac{0.059060 g}{44 g/mol}=0.0013 mol$

Moles of carbon in 0.0013 moles of carbon dioxide gas = 1 × 0.0013 mol = 0.0013 mol

Mass of 0.0013 moles of carbon = $12 g/mol\times 0.0013 mol=0.0156 g$

Mass of water = 24.176 mg = 0.024176

Moles of water = $\frac{0.024176 g}{18 g/mol}=0.0013 mol$

Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol

Mass of 0.0013 moles of hydrogen= $1 g/mol\times 0.0013 mol=0.0013 g$

Mass of sulfur dioxide = 20.326 mg = 0.020326 g

Moles of sulfur dioxide = $\frac{0.020326 g}{64 g/mol}=0.00032 mol$

Moles of sulfur in 0.00032 moles of sulfur dioxide = 1 × 0.00032 mol = 0.00032 mol

Mass of 0.00032 moles of sulfur = $32 g/mol\times 0.00032 mol=0.01024 g$

Mass of oxygen in the sample = x

Mass of sample = 33.153 mg = 0.033153 g

0.033153 g = 0.0156 g + 0.0013 g + 0.01024 g + x

x = 0.006013 g

Moles of oxygen = $\frac{0.006013 g}{16 g/mol}=0.00038 mol$

For empirical formula divide the lowest number of moles of elemnt from all the moles of the all the elements:

Carbon : $\frac{0.0013 mol}{0.00032 mol}=4$

Hydrogen: $\frac{0.0026 mol}{0.00032 mol}=8$

Sulfur : $\frac{0.00032 mol}{0.00032 mol}=1$

Oxygen : $\frac{0.00038 mol}{0.00032 mol}=1$

The empirical formula of the compound = $C_4H_8S_1O_1$

8 0

4 years ago

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

Ahat [919]

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to determine the oxidation state of the elements in the compound.

(a) $H_2SO_4$

Let the oxidation state of 'S' be, 'x'

$2(+1)+x+4(-2)=0\\\\x=+6$

Hence, the oxidation state of 'S' is, (+6)

(b) $Ca(OH)_2$

Let the oxidation state of 'Ca' be, 'x'

$x+2(-2+1)=0\\\\x=+2$

Hence, the oxidation state of 'Ca' is, (+2)

(c) $BrOH$

Let the oxidation state of 'Br' be, 'x'

$x+(-2)+1=0\\\\x=+1$

Hence, the oxidation state of 'Br' is, (+1)

(d) $ClNO_2$

Let the oxidation state of 'N' be, 'x'

$-1+x+2(-2)=0\\\\x=+5$

Hence, the oxidation state of 'N' is, (+5)

(e) $TiCl_4$

Let the oxidation state of 'Ti' be, 'x'

$x+4(-1)=0\\\\x=+4$

Hence, the oxidation state of 'Ti' is, (+4)

(f) $NaH$

Let the oxidation state of 'Na' be, 'x'

$x+(-1)=0\\\\x=+1$

Hence, the oxidation state of 'Na' is, (+1)

4 0

3 years ago

Complete this nuclear reaction by selecting which particle would go in the

kompoz [17]

Answer:

Missing particles: $^{0}_{1}e^{+}$ (a positron) and an electron neutrino $\nu_{\rm e}$ .

The nuclear equation would be:

${\rm ^{19}_{10} Ne} \to ^{0}_{1}e^{+} + {\rm ^{19}_{\phantom{1}9}F }+ \nu_{e}$ .

Explanation:

The mass number of a particle is the number on the top-right corner of its symbol.

The atomic number of a particle is the number on the lower-right corner of its symbol.

The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by $1$ .

A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron $^{0}_{1}e^{+}$ and an electron neutrino $\nu_e$ would be released. On the other hand, in a beta-minus decay, a negatively-charged electron $\rm ^{0}_{1}e^{-}$ and an electron antineutrino $\overline{\nu}_e$ would be released.

Electric charge needs to be conserved in nuclear reactions, including this one.

The atomic number of the $\rm Ne$ nucleus on the left-hand side is $10$ , meaning that the nucleus has a charge of $+10$ . On the other hand, the atomic number of the $\rm F$ nucleus on the right-hand side shows that this nucleus carries a charge of only $+9$ .

By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of $+1$ . That rules out the possibility of the combination of one negatively-charged electron $\rm ^{0}_{1}e^{-}$ (with a charge of $-1$ ) and an electron antineutrino $\overline{\nu}_e$ (with no electric charge at all.)

Hence, the only possibility is that the missing particle is a positron (and an electron neutrino $\nu_e$ , which carries no electric charge.)

4 0

3 years ago

Why is the time series useful as a management tool?

katrin [286]

Answer:

A)

Explanation:

It estimates the probability of extinction.

8 0

3 years ago

Lons form because the loss of one or more electrons makes

ruslelena [56]

This statement lons form because the loss of one or more electrons makes them less stable than their corresponding atoms; this state requires more energy to maintain is FALSE.

An ion forms when an atom loses or gains one more electrons.
This loss or gain of electrons causes the ion to gain the stable octet configuration.

This stable octet configuration requires less energy to maintain than the corresponding atom.

So, since the stable octet configuration of the ion requires less energy than the corresponding atom, the statement is false.

So, the statement lons form because the loss of one or more electrons makes them less stable than their corresponding atoms; this state requires more energy to maintain is FALSE.

Learn more about ions here:

brainly.com/question/3631272

7 0

3 years ago