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2 years ago

5

The ratio of the actual value of a colligative property to the value calculated, assuming the substance to be a nonelectrolyte,

is referred to as ________.

1 answer:

vova2212 [387]2 years ago

5 0

Answer:

It is referred to as the van't Hoff factor.

Explanation:

The van't Hoff factor is named after the Dutch chemist Jacobus Henricus van 't Hoff. It can also be defined as the ratio of the actual quantity of particles to the quantity of particles for no ionization.The fundamental assumption of the van't Hoff factor is that the substance is a nonelectrolyte.

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A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr

sergey [27]

Answer:

184.62 ml

Explanation:

Let $p_1, v_1,$ and $T_1$ be the initial and $p_2, v_2,$ and $T_2$ be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

$p_1=p_2$ ...(i)

$v_1$ = 200 ml

$T_1= 26 ^{\circ}C = 273+26 =299$ K

$T_2= 3 ^{\circ}C = 273+3 =276$ K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

$p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)$

For the final condition,

$p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)$

Equating equation (i), and (ii)

$\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}$

$\frac{v_1}{T_1}=\frac{v_2}{T_2}$ [from equation (i)]

$v_2=\frac{T_2}{T_1} \times v_1$

Putting all the given values, we have

$v_2=\frac{276}{299} \times 200 = 184.62 \; ml$

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

7 0

2 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

8 0

2 years ago

Why aren't flowers sold at a monastery math worksheet answers?

maria [59]

1) x/9 + 16 = 5.

a) x/9 + 16 - 16 = 5 -16; subtraction.

b) x/9 = -11; simplifying.

c) x/9 · 9 = -11 · 9; multiplication.

d) x = -99; simplifying.

2) 5 · (2a - 3) = 60.

a) 10a - 15 = 60; distributive.

b) 10 - 15 + 15 = 60 + 15; addition.

c) 10a = 75; simplifying.

d) 10a ÷ 10 = 75 ÷ 10; commutative.

e) a = 7.5; simplifying.

3) 4 n + (7 + n) = 52.

a) 4n + (n + 7) = 52; commutative.

b) ( 4n + n) + 7 = 52; associative.

c) 5n + 7 = 52; combining.

d) 5n + 7 - 7 = 52 - 7; subtraction.

e) 5n = 45; simplifying.

f) 5n ÷ 5 = 45 ÷ 5; division.

g) n = 9; simplifying.

4) 8t - 3 · (4t + 2) = 16.

a) 8t - 12 t - 6 = 16; distributive.

b) -4 t - 6 = 16; combining.

c) - 4 t - 6 + 6 = 16 + 6; addition.

d) -4t = 22; simplifying.

e) -4t ÷ (-4) = 22 ÷ (-4); division.

f) t = -5.5; simplifying.

This is the question I found on internet.

8 0

3 years ago

Read 2 more answers

The price elasticity of demand measures a. buyers’ responsiveness to a change in the price of a good. b. the extent to which dem

-BARSIC- [3]

Option A

The price elasticity of demand measures buyers’ responsiveness to a change in the price of a good.

<u>Explanation:</u>

Price elasticity of demand holds the responsiveness of need subsequent a variation in a product's cost. In different terms, it’s a process to comprehend out the responsiveness of buyers to inconstancies in cost. Price elasticity estimates the responsiveness of the measure necessitated or outfitted of a good to a shift in its demand.

The price elasticity of demand is the rate fluctuation in the amount demanded of a good or assistance distributed by the percentage shift in the price. Considering the quantity demanded habitually declines with value, the price elasticity coefficient is essentially forever negative.

6 0

2 years ago

[BrO- ]initial is 0.56 M. 33 seconds later [BrO- ] is 0.14 M. What is the rate of change of [BrO- ] in M/s? (The brackets, [ ],

iVinArrow [24]

Answer:

0.013 M/s

Explanation:

Given data

Initial concentration of BrO⁻: 0.56 M

Final concentration of BrO⁻: 0.14 M

Time (t): 33 s

Since the final concentration of BrO⁻ is lower than the initial concentration of BrO⁻, BrO⁻ is dissapearing. The rate of dissapearance of BrO⁻ is:

rBrO⁻ = - Δ [BrO⁻] / t

rBrO⁻ = - (0.14 M - 0.56 M)/ 33 s

rBrO⁻ = 0.013 M/s

3 0

2 years ago

Read 2 more answers