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2 years ago

14

This graph compares the amounts of three different substances that can dissolve in 100 g of water at temperatures between 0° C a

nd 100° C.
Which substance is most soluble?

substance 1

substance 2

substance 3

2 answers:

oksian1 [2.3K]2 years ago

3 0

Substance 2 I think sorry if I’m wrong

bearhunter [10]2 years ago

3 0

I also think it’s substance 2

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Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s

Slav-nsk [51]

Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.

6 0

3 years ago

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb

Bogdan [553]

<u>Answer:</u> The molecular formula for the menthol is $C_{10}H_{20}O$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.0129g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = $\frac{0.0064}{0.00066}=9.69\approx 10$

For Hydrogen = $\frac{0.0129}{0.00064}=19.54\approx 20$

For Oxygen = $\frac{0.00066}{0.00066}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is $C_{10}H_{20}O_1=C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156.27g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Thus, the molecular formula for the menthol is $C_{10}H_{20}O$

4 0

3 years ago

Read 2 more answers

g Which of the following is TRUE for a system that is in dynamic equilibrium? Which of the following is TRUE for a system that i

Firdavs [7]

Answer:

none of the above

Explanation:

A system is said to have attained dynamic equilibrium when the forward and reverse reactions proceed at the same rate. That is;

Rate of forward reaction = Rate of reverse reaction

The implication of this is that the concentrations of reactants and products remain constant when dynamic equilibrium is attained in a system. This does not mean that the reactant and product concentrations become equal; it rather means that their concentrations do not significantly change once dynamic equilibrium has been attained.

3 0

3 years ago

How do the differences in the polyatomic ions po3 3- and po4 3- help you determine whether each ends in -ite or -ate?

Tcecarenko [31]

There is a rule in naming polyatomic anions containing oxygen. Those names ending in <em>–ate</em> are given to the most common oxyanion of an element. While those names ending in <em>–ite</em> are for those that contains fewer O atom for the same charge. Since P $PO_{3} ^{3-}$ has fewer O atoms, its name ends in <em>–ite</em>. Thus, P $PO_{3} ^{3-}$ is phosphite while P $PO_{4} ^{3-}$ is phosphate.

4 0

3 years ago

Determine the oxidation number of each element in these compounds or ions. (a) au2(seo4)3 (gold(iii) selenate) au = se = o = (b)

Stels [109]

<span> Au</span>₂(SeO₄)₃

O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2

Means (SeO₄) contains -2 charge, Now multiply -2 by 3

-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0

Au₂ = +6
Or,
Au = 6 / 2

Au = +3
Result:
Au = +3
Se = +6
O = -2

Ni(CN)₂

Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0

Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2

6 0

3 years ago