Answer:
[He]: 2s² 2p⁵.
[Ne]: 3s².
[Ar]: 4s² 3d¹⁰ 4p².
[Kr]: 5s² 4d¹⁰ 5p⁵.
[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².
Explanation:
- Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.
He contains 2 electrons fill 1s (1s²).
So, [He] can be written before the electronic configuration of 2s² 2p⁵.
Ne contains 10 electrons fill (1s² 2s² 2p⁶).
So, [Ne] can be written before the electronic configuration of 3s².
Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).
So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².
Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).
So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.
Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).
So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².
It’s
1.A
2.C
3.B
hope it’s correct
Answer:
Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)
Explanation:
Use of aqueous sodium hydroxide is a precipitation reaction to test for anions or cations. Aqueous sodium hydroxide in a precipitate test forms a insoluble precipitates along with some colors characteristics.
Aqueous sodium hydroxide (NaOH) when mixed with copper(II) (Cu2+) forms a blue precipitate. The formula is as follows:
Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)
1) Adding salt on the roads in the winter time,
2) Adding salt to the ice bath when making ice cream.
3) Mixing 2 substances when making coolant for a car.
Answer:
0.6 grams of hydrogen are needed to react with 2.75 g of nitrogen.
Explanation:
When hydrogen and nitrogen react they form ammonia.
Chemical equation:
N₂ + 3H₂ → 2NH₃
Given mass of nitrogen = 2.75 g
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2.75 g / 28 g/mol
Number of moles = 0.098 mol
Now we will compare the moles of nitrogen with hydrogen from balance chemical equation:
N₂ : H₂
1 : 3
0.098 : 3×0.098 = 0.3 mol
Mass of hydrogen:
Mass = number of moles × molar mass
Mass = 0.3 mol × 2 g/mol
Mass = 0.6 g
