Answer:
an object is a rest upon a table
The ionic equation will be:
<span>Mg2+(aq) + 2OH-(aq) + 2H+(aq) + 2NO3-(aq) → Mg2+(aq) + 2NO3-(aq) + 2H2O(l)
</span>And the net ionic:
<span>2OH-(aq) + 2H+(aq) → 2H2O(l)
</span>a Balanced equation would be:
<span>Mg(OH)2(aq) + 2HNO3(aq) → Mg(NO3)2 (aq) + 2H2O(l)
</span>Now thsi si so because of the standard equation which is base plus acid=Salt plus water. We need to have in mindo that Mg(OH)2 is sparingly soluble in water and is aqueous mos of the times
Answer:
C. Potassium-19
.
Explanation:
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In this case, since isotopes are known be atoms of the same element with equal atomic number but different mass number, we can consider the case of carbon which has two natural occurring ones, carbon-12 and carbon-13 whereas carbon-12 has the greatest abundance. However, isotope notation may take two forms:
1. Symbol of the element followed by a dash indicating the mass number of the isotope, for instance: C-12, K-39, and so on.
2. Name of the element followed by a dash indicating the mass number of the isotope, for instance: Carbon-12, Potassium-39, and so on.
In such a way, the improper isotope notation is C. Potassium-19
, considering that A should be K-39 because atomic symbol of potassium is K, not k.
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The events that occur is based on the type of hormone being released. Like puberty. It's an event coming from different types of hormones being sent to a tissue
Answer:
24.03 J/mol.ºC
Explanation:
For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.
-Qal = Qw + Qc (minus signal represents that the heat is lost)
-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc
Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC
-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)
1971.15Cal = 1699.10 + 58.22
1971.15Cal = 1757.32
Cal = 0.89 J/g.ºC
The molar mass of Al is 27 g/mol
Cal = 0.89 J/g.ºC * 27 g/mol
Cal = 24.03 J/mol.ºC
