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tekilochka [14]
3 years ago
6

Sugars are polar substances. Which action would make sugar more soluble in water?

Chemistry
2 answers:
azamat3 years ago
5 0

Answer: Option (A) is the correct answer.

Explanation:

When we increase he temperature then the molecules of sugar ans water will start to vibrate with more energy.

As a result, sugar molecules will spread widely into the water and thus, they will start to dissolve. So, sugar molecules being polar in nature will dissolve easily in polar solve (water) as like dissolve like.

Thus, we can conclude that the action of increasing the water temperature would make sugar more soluble in water.

valentinak56 [21]3 years ago
5 0

A.  increase the temperature of the water the higher the temp the more soluble

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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
4 years ago
Read 2 more answers
A chemist dissolves 240mg of pure barium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (Th
goblinko [34]

Answer:

pH = 12.22

Explanation:

<em>... To make up 170mL of solution... The temperature is 25°C...</em>

<em />

The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:

Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)

<em>Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.</em>

<em />

To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:

<em>Moles Ba(OH)₂ molar mass: 171.34g/mol</em>

0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =

2.80x10⁻³ moles of OH⁻

<em>Molarity [OH⁻] and [H⁺]</em>

2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M

As Kw at 25°C is 1x10⁻¹⁴:

Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]

[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M

<em>pH:</em>

pH = -log [H⁺]

pH = -log [6.068x10⁻¹³M]

<h3>pH = 12.22</h3>
8 0
3 years ago
What is helpful about UV rays?
grigory [225]
UV rays can be found In the sun and in tanning beds.
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8 0
3 years ago
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I can’t read ? This what does it say
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3 years ago
What is the concentration of a solution that has 15.0 g NaCl dissolved to a total of 750 ml?
Oksana_A [137]

<u>Answer:</u> The concentration of solution is 0.342 M

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (Sodium chloride) = 15 g

Molar mass of sodium chloride = 58.5 g/mol

Volume of solution = 750 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{15g\times 1000}{58.5g/mol\times 750mL}\\\\\text{Molarity of solution}=0.342M

Hence, the concentration of solution is 0.342 M

6 0
3 years ago
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