-Just look up “H2O lewis structure
-1.5
-Don’t know the VSEPR
-Polar Covalent
-Again, don’t know VSEPR
-Just look up H2O molecule
Answer : The molar mass of the solute will be
87.90 g/mol.Explanation : We know the formula for elevation in boiling point, which is
Δt = i

m
given that, Δt = 0.357,

= 5.02 and mass of

= 40,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X

.
Now, 0.250/ 2.844 X

=
87.90 g/mol. which is the weight of unknown component.
Answer:
3.68 grams.
Explanation:
First we <u>convert 9.5 g of NaCl into moles of NaCl</u>, using its<em> molar mass</em>:
9.5 g ÷ 58.44 g/mol = 0.16 mol NaCl
In<em> 0.16 moles of NaCl there are 0.16 moles of sodium </em>as well.
We now <u>convert 0.16 moles of sodium into grams</u>, using <em>sodium's molar mass</em>:
0.16 mol * 23 g/mol = 3.68 g
Answer:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
CLOUDS would be one of the interacting parts in a weather system.