Answer:
The mixture contains 8.23 g of Ar
Explanation:
Let's solve this with the Ideal Gases Law
Total pressure of a mixture = (Total moles . R . T) / V
We convert T° from °C to K → 85°C + 273 = 358K
3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L
(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles
0.756= Total moles from the mixture
Moles of Ar + Moles of H₂ = 0.756 moles
Moles of Ar + 1.10 g / 2g/mol = 0.756 moles
Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206
We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g
Answer: option C. Copper (II) chloride
Explanation:
To name CuCl2, we need to know the oxidation state of Cu in the compound as chlorine always have oxidation on —1 in all its compound. The oxidation state of Cu can be calculated as follows:
Cu + 2Cl = 0 (since the compound has no charge)
Cl = —1
Cu + 2(—1) = 0
Cu —2 = 0
Collect like terms
Cu = 0 +2
Cu = +2
Therefore, the oxidation state of Cu in CuCl2 is +2.
The name of the compound will be copper(ii) chloride, since cupper has oxidation state +2 in the compound.
Answer:
If an atom looses all of its electrons then it will become positively charged. It will also turn into an Ion.
Explanation:
Gas.
The sky is not solid, not liquid but of gas. It is also
called the atmosphere, one of the important spheres on earth that influences
the climate –temperature, humidity, precipitation, atmospheric pressure, and
the biotic community as a whole. The atmosphere or the sky is composed of many
layers and is responsible to protecting the living organisms of the earth
against the deadly ultraviolet rays and other strong radiation coming from the
sun.
Answer : The concentration of the NaOH solution is, 0.738 M
Explanation :
To calculate the concentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


Thus, the concentration of the NaOH solution is, 0.738 M