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kvv77 [185]
3 years ago
12

Every Sunday, Sarah prepares blue fruit punch to

Chemistry
2 answers:
MrRissso [65]3 years ago
8 0

Answer:

Fruit punch with a very dark blue color

Explanation:

Juli2301 [7.4K]3 years ago
5 0
<h2>Answer: Its Fruit Punch with a very dark blue color</h2>

<h2>Explanation: Just answered it on Edge and got it right.</h2>

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What elements are highlighted:<br> a. Nonmetals<br> b. Metals<br> C. Metalloids
Advocard [28]
Non metals
(It’s both halogens and noble gases ) the other ones which are white(not highlighted) are alkali metals, alkaline earth metals, and transition metals :3
6 0
3 years ago
What is the difference between diffusion and osmoses<br><br>​
Citrus2011 [14]

Answer:

Diffusion means that particles travel from a higher concentration to a lower concentration region before balance is achieved. The semi-permeable membrane is present in osmosis, so that only solvent molecules can travel freely to equalize the levels.

5 0
2 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

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

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

8 0
3 years ago
Can someone please answer this?
Amanda [17]

Answer:-

0.229 L

Explanation:-

Molar mass of AgBr = 107.87 x 1 + 79.9 x 1

=187.77 grams mol-1

Mass of AgBr = 150 grams

Number of moles of AgBr = 150 grams / 187.77 gram mol-1

= 0.8 mol

The balanced chemical equation is

NaBr (aq) + AgNO3 (aq)--> AgBr(s) + NaNO3(aq)

From the equation we can see that

1 mol of AgBr is produced from 1 mol of AgNO3.

∴ 0.8 mol of AgBr is produced from 1 x 0.8 / 1 = 0.8 mol of AgNO3.

Strength of AgNO3 = 3.5 M

Volume of AgNO3 required = Number of moles / strength

= 0.8 moles / 3.5

=0.229 L

8 0
3 years ago
If an isotope decays by the process of beta emission, ___.
Kaylis [27]
1. Decreases by 4. (B)
2. The atomic number changes. (B)
3. 56/26 Fe. (C)
4. Potassium-40;t1/2=25 days. (B)
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3 0
3 years ago
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