Answer:
two electrons
Explanation:
Calcium atoms will lose two electrons in order to achieve the noble gas configuration of argon.
Halogens
Explanation:
Halogens are a group of non-metals located in the seventh group on the periodic table. The will only gain one electron during a chemical reaction.
- Halogens have a seven electrons in their outermost shell.
- To complete the number of electrons in this shell, they need to gain an additional electron.
- One more electron makes the halogen similar to the corresponding noble gas which is very stable.
- Halogens are very reactive groups of elements and are highly electronegative.
- They have a high affinity for electrons.
- These elements are fluorine, chlorine, bromine, iodine and Astatine.
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Answer:
1.7x10^8 Hz
Explanation:
Frequency could be explained as the number of occurrences of a repeating event at a time
Given:
wavelength = 1.8 meters
The frequency f of the waves can be calculated using f = c / λ
Where c (m/s) is the speed of the wave
λ (m) is the wavelength
Speed c= 3*10^8 m/s
Frequency f= 3*10^8 /1.8
Frequency= 1.7x10^8 Hz
Therefore,the frequency of waves from a radar detector is 1.7x10^8 Hz
Answer:
molality of sodium ions is 1.473 m
Explanation:
Molarity is moles of solute per litre of solution
Molality is moles of solute per kg of solvent.
The volume of solution = 1 L
The mass of solution = volume X density = 1000mL X 1.43 = 1430 grams
The mass of solute = moles X molar mass of sodium phosphate = 0.65X164
mass of solute = 106.6 grams
the mass of solvent = 1430 - 106.6 = 1323.4 grams = 1.3234 Kg
the molality = 
Thus molality of sodium phosphate is 0.491 m
Each sodium phosphate of molecule will give three sodium ions.
Thus molality of sodium ions = 3 X 0.491 = 1.473 m
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
