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Nastasia [14]
3 years ago
6

Which element would have similar properties to Sulfur? Chlorine O a ob Fluorine Oxygen Nitrogen​

Chemistry
1 answer:
NISA [10]3 years ago
6 0

Answer:

sir/ma'am, i know i'm a <u>NOOB</u> at this kind of stuff but i believe it's oxygen.

Explanation:

i can't really remember the answer since it's been a week or so that i done this problem, it's between nitrogen and oxygen, though, i believe it's oxygen.

if ya don't believe me or trust me, then here's the link to help ya:

https://socratic.org/questions/what-elements-would-have-similar-properties-to-sulfur#628367

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A cylinder containing carbon dioxide of volume 20 L at 2.0 atm was connected to another cylinder of certain volume at constant t
den301095 [7]

Answer:

The volume of the second cylinder is 80 liters

Explanation:

We use the Boyle-Mariotte formula, according to which the pressure and volume of a gas are inversely related, keeping the temperature constant: P1 x V1 = P2xV2. We convert the pressure in mmHg to atm:

760 mmHg-----1 atm

380mmHg------x= (380mmHgx1atm)/760mmHg=0,5 atm

P1xV1=P2xV2

2 atmx20 L= 0,5atm x V2 V2=(2 atmx20 L)/0,5atm=80L

8 0
3 years ago
2. How many grams of water can be heated from 20.0 oC to 75oC using 12500.0 Joules?
kenny6666 [7]

Answer;

  = 0.054 kg or 54 g

Explanation;

Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.  

ΔT = 75 - 20 = 55 C.  

Solve the equation for m  

m = Q/ cΔT

Mass = 12500 / (55 × 4200)

        = 0.054 kg or 54 g


4 0
3 years ago
Read 2 more answers
. The freezing point of an aqueous solution containing a nonelectrolyte solute is – 2.79 °C. What is the boiling point of this s
Semmy [17]

Answer:

Boiling point of the solution is 100.78°C

Explanation:

This is about colligative properties.

First of all, we need to calculate molality from the freezing point depression.

ΔT = Kf . m . i

As the solute is nonelectrolyte, i = 1

0°C - (-2.79°C) = 1.86 °C/m . m . 1

2.79°C / 1.86 m/°C = 1.5 m

Now, we go to the boiling point elevation

ΔT = Kb . m . i

Final T° - 100°C  =  0.52 °C/m . 1.5m . 1

Final T° =  0.52 °C/m . 1.5m . 1  + 100°C → 100.78°C

4 0
3 years ago
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Answer:

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