Answer: 1.414x10^24 molecules in 94.4g MgO
Explanation: molar mass MgO 40.204
molecules in 40.204 g MgO = avogadro number
molecules in 94.4 g MgO = (94.4/40.204)*avogadro number
(94.4/40.204)*6.02214076*10^23 = 14.14x10^23
Answer:
ΔHrxn = 239 kj/mol
Explanation:
Given data:
Mass of sodium = 0.230 g
Heat produced = 2390 J
Solution:
Chemical equation:
2Na + 2HCl → 2NaCl + H₂
Number of moles of sodium:
Number of moles of sodium = mass / molar mass
Number of moles of sodium = 0.230 g / 23 g/mol
Number of moles of sodium = 0.01 mol
Enthalpy of reaction:
ΔHrxn = 2390 J / 0.01 mol
ΔHrxn = 239000 j/mol
ΔHrxn = 239 kj/mol
The radius of the football would be 748299320 times bigger than that of one atom of fluorine.
<h3>Atomic radius of fluorine</h3>
The atomic radius of a fluorine atom is 147 picometers or 147 x m.
The radius ratio of a football with 11 cm radius and that of one atom of fluorine can be calculated as:
11 x /147 x
= 748299320
Thus, the radius of the football would be 748299320 times bigger than that of one atom of fluorine.
More on the atomic radius of fluorine can be found here: brainly.com/question/10330465
(On my left) (On my right side)
Na: 1. Na:1
Cl:2. Cl:1
Na:1 x 2 =2
Na:1 x 2=2
Cl: 1 x 2 = 2
Answer:
2Na + Cl2 -> 2NaCl
Answer: B, C, D, F
Questioning, creative, open-minded, objective
Explanation: