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ololo11 [35]
2 years ago
8

Elements are one type of pure substance. In an element, every atom is the same as every other atom. It

Chemistry
1 answer:
tankabanditka [31]2 years ago
5 0
The answer is B. Hope this helps
You might be interested in
1. What Volume of HCl is generated if 3.44 g of Cl2 are reacted at STP? 2. What volume does 4.87 mol of Kr have at STP? 3. What
IgorLugansk [536]

Answer:

1. 2.17 dm3 VOLUME OF HCl IS PRODUCED WHEN 3.44 g OF Cl2 REACT AT STP

2. 4.87 MOLE OF Kr AT STP CONTAINS 109.088 dm3 .

3. THE PRESSURE WHEN 3.44 g OF Cl2 are reacted at 4.55 L AT 455 K IS 0.77 atm

Explanation:

1 Volume of HCl if 3.44 g of Cl2 are reacted at STP?

Equation for the reaction:

H2 + Cl2 ---------> 2HCl

1 mole of Cl2 reacts to form 2 mole of HCl

At STP, 1 mole of a gas is equal to the molar mass of the gas sample

35.5 * 2 g of Cl2 reacts to form 2 mole of HCl

3.44 g of Cl2 will react to form  ( 3.44 * 2 / 71 ) mole of HCl

= 0.0969 mole of HCl

1 mole of HCl = 22.4 dm3

0.0969 mole of HCl = ( 22.4 * 0.0969 / 1)

= 2.17056 dm3

The volume of HCl is 2.17 dm3 when 3.44 g of Cl2 are reacted at STP.

2. What volume does 4.87 mol of Kr have at STP?

1 mole of a substance is 22.4 dm3 of the sample

1 mole of Kr = 22.4 dm3

4.87 mole of  Kr = 4.87 * 22.4

= 109.088 dm3

4.87 mole of Kr at STP contains 109.088 dm3 volume

3. Whta pressure of HCl is generated if 3.44 g of Cl2 are reacted at 4.55 L at 455 K

Using the formula:

PV = nRT

V = 4.55 L

R = 0.082 L atm/ mol K

T = 455 K

m = 3.44 g

n = mass / molar mass

molar mass = ( 1 + 35.5) = 36.5 g/mol

n = 3.44 g / 36.5 g/mol

n = 0.094 mole

P = nRT / V

P = 0.094 * 0.082 * 455 / 4.55

P = 3.50714 / 4.55

P = 0.7708 atm

The pressure of HCl if 3.44 g of Cl2 are reacted at 4.55 L and 455 K is 0.7708 atm.

8 0
3 years ago
In a
dsp73

Answer:

1384 kJ/mol

Explanation:

The heat absorbed by the calorimeter is equal to the heat released due to the combustion of the organic compound. C is the total heat capacity of the calorimeter and Δt is the change in temperature from intial to final:

Q = CΔt = (3576 J°C⁻¹)(30.589°C - 25.000°C) = 19986.264 J

Extra significant figures are kept to avoid round-off errors.

We then calculate the moles of the organic compound:

(0.6654 g)(mol/46.07) = 0.0144432 mol

We then calculate the heat released per mole and convert to the proper units. (The conversion between kJ and J is infinitely precise and is not involved in the consideration of significant figures)

(19986.264 J)(1kJ/1000J) / (0.0144432 mol) = 1384 kJ/mol

8 0
3 years ago
10. On the basis of the general solubility rules given
Crank
I think it’s c but I could be wrong
3 0
3 years ago
What are the requirements for combustion of a candle?
cestrela7 [59]
For a candle to burn, it requires a spark, which provides the activation energy for the oxidation reaction of the hydrocarbon making the candle.
It also requires oxygen to facilitate the oxidation of the hydrocarbon.
Therefore the two main requirements of combustion of a candle are oxygen and a spark (or an initial flame)
7 0
3 years ago
This element exists in adundance in the sun.Explain how you would go about capturing sunlight.Would this captured sunlight conta
nasty-shy [4]
Yes because the sun will make alots of thing grow
4 0
3 years ago
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