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zavuch27 [327]
4 years ago
13

I JUST GOT RICKROLLED WHAT DO I DO PLEEEAASSEEEE HELP

Chemistry
2 answers:
soldi70 [24.7K]4 years ago
8 0

Answer:

contemplate your life...what you are doing here...what it means to live in our society...THEN...pass it on :0

Explanation:

mind blown ~\_(* - *)_/~

WINSTONCH [101]4 years ago
4 0

Answer:

Give up

Explanation:

You just got rickrolled, just sit down and think of how your life lead up to that moment, and just think "Never Gonna give you up, Never Gonna let you down, Never gonna run around and desert you, Never gonna make you cry, Never Gonna say goodbye, Never gonna tell a lie and hurt you".

You just got rickrolled.

Sorry had to do it.

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Explain how natural resources are identified and why natural resources are unevenly distributed.
slava [35]

Answer:

Natural resources are naturally formed of course(wood copper fruits) and they are not distributed evenly due to competition of getting them and the prices of the resources

Explanation:

5 0
3 years ago
If an insect population increases and then decreases as the available supply of food changes, what does this demonstrate
Semenov [28]
D.

This is self-regulation because when the population of the insects becomes too large, it regulates itself and starts to decrease due to a shortage of resources.
7 0
4 years ago
Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

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

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\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 pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

3 0
3 years ago
If the pressure inside the cylinder increases to 1.3 atm, what is the final
EleoNora [17]

Answer:

1.4 × 10² mL

Explanation:

There is some info missing. I looked at the question online.

<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>

Step 1: Given data

  • Initial volume (V₁): 215 mL
  • Initial pressure (P₁): 625 mmHg
  • Final volume (V₂): ?
  • Final pressure (P₂): 1.3 atm

Step 2: Convert 625 mmHg to atm

We will use the conversion factor 1 atm = 760 mmHg.

625 mmHg × 1 atm/760 mmHg = 0.822 atm

Step 3: Calculate the final volume of the air

Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL

5 0
3 years ago
The store paid 4.50 for a book and sold it for $7.65.what is the profit as a percent of the cost to the store? A. 3. 15% b. 58.8
xeze [42]

Answer: This would be C.) 70%

Explanation:

4 0
3 years ago
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