The component of the candle burning in the surrounding has been the oxygen in the air.
The burning of candle wax and wick has been the chemical reaction. It has been based on the reaction of wick with the atmospheric oxygen, resulting in the formulation of the wax burning.
<h3>Chemical reaction of burning of wax</h3>
The wax has been vaporizes by the heat of the flame, that has been resulted by the burning. The wick has been able to react with the oxygen and form the byproducts that helps in flame burning.
The end products have been wick and oxygen as the wax has been consumed in the reaction. The air in the surrounding has oxygen as the part of the system, as it has been involved in the reaction.
Learn more about candle burning, here:
brainly.com/question/25955977
Answer:
The nuclear charge increases from boron to carbon, but there is no additional shielding( that is no additional shells).
Explanation:
First of all, we must know the electron configuration of carbon and boron.
Boron- 1s2 2s2 2p1
Carbon- 1s2 2s2 2p2
Moving from boron to carbon, the effective nuclear charge increases without a corresponding increase in the number of shells. Remember that shielding increases with increase in the number of intervening shells between the outermost electron and the nucleus. Since there isn't an increase in shells, boron experience a lower screening effect.
From
Zeff= Z- S
The Z for carbon is 6 while for boron is 5 even though both have the same number of screening electron S(4 screening electrons). Hence it is expected the Zeff(effective nuclear charge) for boron will be less than that of carbon.
7 becuse it splits in half
Answer:
V = 34430 mL
Explanation:
Given data:
Volume in mL = ?
Number of moles of gas = 2.00 mol
Temperature = 36°C (36+273= 309K)
Pressure of gas = 1120 torr
Solution:
Formula:
PV = nRT
V = nRT/P
V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr
V = 38563.2 torr • L / 1120 torr
V = 34.43 L
L to mL
34.43 L ×1000 mL / 1 L
34430 mL
Molar solubility is number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.
The molar solubility of lead(ii) chloride with ksp value of 2.4 × 10e4 can be solve as:
Ksp = s2 = 2.4 × 10e4
s2 = 2.4 × 10e4
s = √(2.4 × 10e4)
s = 154.9 mol/L
