Answer 1) Option D) 3 moles.
Explanation : In the reaction we can see that the number of moles of hydrogen are equal to the number of water molecules formed. Also, oxygen is found t o be the limiting reactant as equal number of moles are not produced in the reaction.
Here when we use 3 moles of hydrogen then 3 moles of water will be produced. As we are ignoring the moles of oxygen here as that is a limiting reactant.
Answer 2) Option A) 9 moles of reactants chemically change in to 11 moles of product.
Explanation : We can see that 1 mole of pentane reacts with 8 moles of oxygen to produce 5 moles of carbon dioxide and 6 moles of water. So when we add the reactants side we get 1 + 8 = 9 moles and in products it has 5 + 6 = 11 moles.
Hence, 9 moles of reactants produce 11 moles of products.
Answer 3) Option A) 0.62 moles.
Explanation : Here, the number of moles of Cu reacts with 2 moles of silver nitrate and gives 1 mole of copper nitrate and 2 moles of silver.
So, the ratio is 1 : 2 in the reactants side, Now, when we have 1.23 moles of silver nitrate then to calculate the number of Cu.
we can simply cross multiply and get the answer as 1.23 / 2 = 0.615 = ~ 0.62 moles.
Answer:
d. a brittle solid.
Explanation:
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In this case, since cesium fluoride is widely known as an ionic compound due to the large electronegativity difference, it is possible to discard a. due to its crystalline structure, b. because conducting solids are metals, c. because ionic compounds are not likely to be gases and e. because even when it is soluble in water, the problem is not referring to an aqueous solution; therefore, the answer is d. a brittle solid.
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Answer:
You find the molar mass of calcium metal, it is listed as 40.1⋅g⋅mol−1 . You should check your copy of the Periodic Table to see if I have got it right. This means that Avogadro's number of calcium atoms, i.e. 6.022×1023 individual calcium atoms have a mass of 40.1⋅g .
Explanation:
a) Alkali metals
=> group 1
=> Li: 1s2 2s => 1s
Na: [Ne] 3s => 3s
K: [Ar] 4s => 4s
Rb: [Kr] 5s => 5s
Cs: [Xe] 6s => 6s
Fr: [Rn] 7s => 7s
=> outer electron configuration is ns, where n is the main energy level: 1, 2, 3, 4, 5, 6,7.
b) Alkaline earth metals
=> group 2 => you have to add 1 electron to the alkaly metal of the same row.
=> Be: [He] 2s2 => 2s2
Mg: [Ne] 3s2 => 3s2
Ca: [Ar] 4s2 => 4s2
Sr: [Kr] 5s2 => 5s2
Ba: [Xe] 6s2 => 6s2
Ra: [Rn[ 7s2 => 7s2
=>outer electron configuration is n s2, where n is the main energy level: 1, 2, 3, 4, 5, 6, 7
c) halogens
=> group 7
=> F: [He] 2s2 2p5 => 2s2 2p5
Cl: [Ne] 3s2 3p5 => 3s2 3p5
Br: [Ar] 3d10 4s2 4p5 => 4s2 4p5
I: [Kr] 4d10 5s2 5p6 => 5s2 5p5
At: [Xe] 4f14 5d10 6s2 6p5
=> outer electron configuration is ns2 np5, where n is the main energy level 1, 2, 3, 4, 5, 6, 7
d) Noble gases
=> group 8
I will show only the outer shell which is what is requested
=> He: 1s2
Ne: ... 2s2 2p6
Ar: ... 3s2 3p6
Kr: ... 4s2 4p6
Xe: ... 5s2 5p6
Rn: ... 6s2 6p6
=> the outer electron configuration is ns2 np6, except for He for which it is 1s2