Answer:
17.04 g/mol
Explanation:
Molar Mass of NH₃
we know that
Nitrogen has 14.01 gram/mol
And Hydrogen has 1.01 gram/mol
but we have 3 Hydrogens So we multiply
1.01 by 3 i.e., 3.03
Now, add
14.01
+<u> </u><u>3</u><u>.</u><u>0</u><u>3</u>
17.04
So, The molar mass of ammonia, NH₃ is
17.04 g/mol
<u>-TheUnknown</u><u>Scientist</u>
The ions of Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
The periodic table is a systematic arrangement of elements in order of their atomic numbers into a set of 8 columns each called groups and a set of 7 rows each called a period.
Elements are arranged in different groups according to the number of Valence electrons they have.
- For instance, elements in the group I of the periodic table are highly electropositive and as such are highly reactive.
The same is evident in group 7 elements are highly electronegative and have high electron affinity and as such are unstable and reactive.
- However, Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
Consequently, the <em>Noble gases ion</em> has a stable Valence electron configuration.
Read more:
brainly.com/question/5336231
First, we have to see how K2O behaves when it is dissolved in water:
K2O + H20 = 2 KOH
According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:
c(KOH) : c(K2O) = 1 : 2
So,
c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)
Now we can calculate pH:
pOH= -log c(OH⁻) = -log 0.01 = 2
pH= 14-2 = 12
In the reaction Sn(s) + 2H+(aq) → Sn2+ (aq) + H2(g)
from this reaction, we get that Sn loses from 0 to 2 electrons so it's oxidized So it is the reducing agent.
and H gains from 0 to 1 electrons so, it's reduced so ∴ it is the oxidizing agent
Boiling point<span> is the </span>temperature<span> at which the vapor pressure of the liquid equals the surrounding pressure.
Above boiling point point, liquid get converted into vapour.
Now, boiling point of water is 100 oC at room pressure. Room pressure is equal to 760 torr. Thus, at 100 oC, vapour pressure of water becomes equal to 760 torr.
Now, if external pressure is increased to 880 torr, more heat is to be supplied so that vapour pressure of water equals 880 torr.
So, at 880 torr, boiling point of water will be more than 100 oC. In present case, most like the boiling point of water is equal to 105 oC.
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