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2 years ago

State whether each of the following will be more soluble in water or hexane i. Butane ii. Ch3cooh iii. K2so4

Chemistry

1 answer:

ioda2 years ago

3 0

Explanation:

Solubility is determined by the principle , "like dissolves like" .

i.e. , if a compound is polar then it will dissolve in a polar compound only , and

if a compound is non - polar then it will dissolve in a non - polar compound only .

Hence , from the question ,

Water is a polar molecule , and hence it will dissolve only the polar molecule , i.e. , from the given options the polar molecule is , iii. K₂SO₄

Hexane , is a non - polar molecules , hence it will dissolve only the non polar molecule , i.e. , from the given options the non polar molecule is i. Butane .

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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago

A fictional element has two isotopes and an atomic mass of 87.08 amu. if the first isotope is 86 amu and the second isotope has

Lorico [155]

The atomic mass of a certain element is summation of the product of the decimal equivalent of the percentage abundance and the given atomic mass of each of the isotope. If we let x be the percentage abundance of the 86 amu-isotope then, the second one is 1-x such that,
x(86) + (1 - x)(90) = 87.08
The value of x from the equation is 0.73. This value is already greater than 0.5. Thus, the isotope with greatest abundance is that which is 86 amu.

4 0

2 years ago

If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?

kap26 [50]

Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3 1mol CO2
---------------------- x -------------------------
1 84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!

6 0

2 years ago

Read 2 more answers

Explain why groups 14 and 15 are better representatives of mixed groups than groups 13 and 16

Brrunno [24]

<h3><u>Answer</u>;</h3>

Groups 14 and 15 each contain metals, nonmetals, and metalloids while Group 13 contains metals and a metalloid, and Group 16 contains metalloids and nonmetals.

<h3><u>Explanation;</u></h3>

Groups 13–16 of the periodic table contain one or more metalloids, in addition to metals, nonmetals, or both.
Unlike other groups of the periodic table, which contain elements in one class, groups 13–16 referred to as mixed groups contain elements in at least two different classes. In addition to metalloids, they also contain metals, nonmetals, or both.
<em><u>Group 14 also known as the carbon group contains carbon which is a non metal, silicon and germanium which are metalloids and tin and lead which are metals.</u></em>
<em><u>Group 15 also known as the Nitrogen group contains non metals such as oxygen, metalloid tellurium and a metal polonium.</u></em>

3 0

2 years ago

For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o

galina1969 [7]

Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

S S 2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

⇒ S³ = 1.6 E-9

⇒ S = 1.1696 E-3 M

NaF(s) → Na+(aq) + F-(aq)

0.10M 0.10M 0.10M

MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

S' S' 2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

8 0

2 years ago