Answer is: <span>C.)It will be closer to the mass of Vi–193.
</span><span>Average atomic mass of vitellium =
Ar(Vi-188) </span>· ω(Vi-188) + Ar(Vi-191) · ω(Vi-191) + Ar(Vu-193) · ω(Vi-193).
Average atomic mass of vitellium =
187.9122 amu · 0,10861 + 190.9047 amu · 0,12428 + 192.8938 amu · 0,76711.
Average atomic mass of vitellium = 20,409 + 23,725 + 147,970.
Average atomic mass of vitellium = 192,104.
Answer:
1. Polar
2 nonpolar
3. Nitrogen or oxygen
4.carbon
5.Repelled by and attracted to
It looks like we are solving for a pressure. All that is required is some algebraic manipulation to find our pressure in mmHg.
Given:
(5.0 m³)(7.5 mmHg) = (P)(4.0m³)
Multiply:
37.5 = 4.0P
Divide:
9.375 = P
P = 9.4 mmHg (remember sig figs)
<h3>
Answer:</h3>
9.4 mmHg
Answer:
<h2>pH = 3.9</h2><h2>pOH = 10.1</h2>
Explanation:
Since
is a weak acid to find the pH of
we use the formula

where
Ka is the acid dissociation constant
c is the concentration
From the question
Ka of
= 1.75 × 10^-5
c = 1.00 × 10-³M
Substitute the values into the above formula and solve for the pH
That's

We have the answer as
<h3>pH = 3.9</h3>
To find the pOH we use the formula
pH + pOH = 14
pOH = 14 - pH
pOH = 14 - 3.9
We have the answer as
<h3>pOH = 10.1</h3>
Hope this helps you
Answer:
(molecular) 3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)
(ionic) 3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)
(net ionic) 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)
Explanation:
The molecular equation includes al the species in the molecular form.
3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)
The ionic equation includes all the ions (species that dissociate in water) and the species that do not dissociate in water.
3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the species that do not dissociate in water. In does not include <em>spectator ions</em>.
3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)