Which climate prevails in the northwest coast of the United States?

Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling

Reil [10]

Answer:

The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is 394.91 K.

Explanation:

The formula for molal boiling Point elevation is :

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b}$ = elevation in boiling Point

$K_{b}$ = Boiling point constant( ebullioscopic constant)

m = molality of the solution

i = Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

$Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}$

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

i = Van't Hoff Factor = 5

m = 8.5 m

$K_{b}$ = 0.512 °C/m

Insert the values and calculate temperature change:

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b} = 5\times 0.512\times 8.5$

$\Delta T_{b} = 21.76 K$

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

$\Delta T_{b} = T_{b} - T_{b}_{pure}$

$T_{b}_{pure}$ = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

8 0

2 years ago

Determain the number of moles in 2.24l of ch4 at stp

valkas [14]

Answer:

0.1 mole of CH₄

Explanation:

From the question given above, the following data were obtained:

Volume of CH₄ = 2.24 L

Number of mole of CH₄ =?

The number of mole of CH₄ can be obtained as follow:

Recall:

1 mole of a gas occupy 22.4 L at stp. This implies that 1 mole of CH₄ occupies 22.4 L at stp.

22.4 L = 1 mole of CH₄

Therefore,

2.24 L = 2.24 × 1 mole of CH₄ / 22.4

2.24 L = 0.1 mole of CH₄.

6 0

2 years ago

In a chemical reaction, what is defined as the difference between the potential energy of the products and the potential energy

zlopas [31]

I Believe That The Answer Is A
Hope This Helps !

6 0

2 years ago

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Two samples of a compound containing elements a and b are decomposed. the first sample produces 15 g of a and 35 g of

Alborosie

According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.

The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.

Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases

\frac{a}{b}=\frac{15}{35}=\frac{10}{y}

rearranging,

y=\frac{10\times 35}{15}=23.3

Thus, mass of b produced will be 23.3 g.

3 0

3 years ago

Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha

Aleks04 [339]

Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

$pH = pKa + log \frac{[Base]}{[Acid]}$

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

$7.5 = 6.3 + log \frac{[base]}{[acid]}$

$1.2 = log \frac{[base]}{[acid]}$

$\frac{[Base]}{[Acid]} = 10^{1.2}$

$\frac{[Base]}{[Acid]} = 15.8$

$[Base] = 15.8 \times [Acid]$

The total of mole fraction of acid and base is 1. Therefore we have,

$[Acid] + [Base] = 1$

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

$[Acid] + 15.8 \times [Acid] = 1$

$16.8 [Acid] = 1$

$[Acid] = \frac{1}{16.8}$

$[Acid] = 0.0595$

[Acid] = 0.0595 x 100 = 5.95 %

The fraction of carbonic acid present in the blood is 5.95%

4 0

3 years ago

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