Answer:
18.73× 10²³ formula units
Explanation:
Given data:
Number of moles of Ca(NO₃)₂ = 3.11 mol
Number of formula units = ?
Solution:
Avogadro number:
"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance"
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
Number of formula units of Ca(NO₃)₂:
1 mole contain 6.022 × 10²³ formula units
3.11 mol × 6.022 × 10²³ formula units / 1 mol
18.73× 10²³ formula units
Answer:
the first one is protons and electrons the second one is protons the third one is neutrons and the forth one is also neutron
Answer:
Explanation:
pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10^−pH,
so PH =2.4 in you case is
[H+] = 10^-2.4 =0.00398
The option are not correct it looks
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
Answer: One formula unit of NaCl consists of one cation, whose chemical symbol is 
and one anion whose chemical symbol is 
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
The cation is formed by the metal sodium which forms and the anion is formed by non metal chlorine which forms .
For a formula unit of sodium chloride, the charges have to be balanced , thus the valencies of ions are exchanged and the neutral compound result. Thus
and combine to form neutral 
