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Katyanochek1 [597]
2 years ago
14

Find the density if the mass is 20 grams and the volume is 10 mL?

Chemistry
1 answer:
tatiyna2 years ago
5 0

Answer:

D= 2g/mL

Explanation:

20g/10ml= 2g/mL

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C12H22O11 is an example of a(n):
ycow [4]

Answer:

molecular compound

Explanation:

it has two or more nonmetal elements and are covalently bonded

5 0
2 years ago
What would the volume of a gas be at 150c if had of volume of 693 ml at 45 c​
Vlad1618 [11]

Answer:

Explanation:

T1 = 150°C = (150 + 273.15)K = 423.15K

T2 = 45°C = (45 + 273.15)K = 318K

V1 = 693mL = 693cm³

Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.

V = kT

V1 / T1 = V2 / T2

693 / 423.15 = V2 / 318

V2 = (693 * 318) / 423.15 = 520.79cm³

The new volume of the gas is 520.79cm³

6 0
3 years ago
Name a molecule that is small enough to enter the cell
Oksanka [162]

Answer:

oxygen molecule

Explanation:

I belive it is oxygen molecule

8 0
2 years ago
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HELP!!! DUE TONIGHT!!
IRINA_888 [86]

Answer:

K I will attempt

Explanation:

a)

O_2_{(g)}+ 2H_2_{(g)} => 2H_2O_{(l)}\\

b)

1 : 2 : 2 (I don't know if this is what the question wants but it is what I would answer)

c)

Hydrogen because it requires 2 moles of H2 to react with 1 mole of O2

d)

24 moles of water. Look at stoichiometric coefficient. 2:2 means 24 moles you get 24 moles

e)

Oxygen. 2 < 5/2. Remember, 1 mole of O2 requires 2 moles of H2. But 5/2 is still greater than 2

f)

First, let's find out how many moles of water we can get. Since O2 is the limiting reactant, and O2:H2O ratio is 1:2, we will get 4 moles of H2O. Then, we can multiply 4 by Avogadro's number which is 6.022*10^{23} to get the number of molecules. We get: 2.41 * 10^24 molecules of water.

6 0
2 years ago
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In a reaction, gaseous reactants form a liquid product. The heat absorbed by the surroundings is 1.1 MJ, and the work done on th
ratelena [41]

Answer: \Delta E is 1155 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=-P\Delta V  {Work done on the system is positive as the final volume is lesser than initial volume}

w =13.2kcal=55.2kJ   (1kcal=4.184 kJ)

q = +1.1 MJ = 1100 kJ  (1MJ=1000kJ)   {Heat absorbed by the system is positive}

\Delta E=1100kJ+(55.2)kJ=1155kJ

Thus \Delta E is 1155 kJ

4 0
3 years ago
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