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Evgen [1.6K]
3 years ago
8

Match the followingOf scinece chapter​

Chemistry
1 answer:
ElenaW [278]3 years ago
8 0

  1. Anthrax - iv. Wool Industry
  2. Mushroom - v. Saprophyte
  3. Trade wind - i. Convection of heat
  4. Chlorophyll - iii. Photosynthesis
  5. Mercury - ii. Thermometric fluid
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Which of the following has the same empirical and molecular formula?
White raven [17]

Answer:

O C4H8

Explanation:

5 0
3 years ago
The compound consists of 40.1% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative density with respect to hydrogen is -15. Wha
kap26 [50]

Answer:

https://socratic.org/answers/220339

this is the answer , my grandpa create this page

Explanation:

3 0
3 years ago
1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?
Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

  • <u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L  

V (final volume) = 473 mL → 0.473 L  

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)  

P (final pressure) = ? (in atm)

  • <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

P_0*V_0 = P*V

1*1 = P*0.473

1 = 0.473\:P

0.473\:P = 1

P = \dfrac{1}{0.473}

\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The new pressure of the gas is 2.11 atm  </em></u>

___________________________________

\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

3 0
3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

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

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
Which of these careers would a student studying biotechnology
matrenka [14]
I think it’s c I could be wrong
3 0
3 years ago
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