Answer:
See explanation
Explanation:
A balanced chemical reaction equation has the same number of atoms of each element on both sides of the reaction equation.
Hence, for the reaction between KOH and H2SO4, the balanced chemical reaction equation is;
H2SO4(aq) + 2KOH(aq) ---------> K2SO4(aq) + 2H2O(l)
Complete ionic equation;
2H^+(aq) + SO4^2-(aq) + 2K^+(aq) +2OH^-(aq) -------> SO4^2-(aq) + 2K^+(aq) + 2H2O(l)
Net ionic equation;
2H^+(aq) + 2OH^-(aq) -------> 2H2O(l)
Answer:
Polyatomic Ionic Compound
Explanation:
In given statement the compound given is called as Sodium oleate this means that when Oleic acid is treated with NaOH then it forms.
In chemistry there are few species which are involved in the formation of compounds.
(i) Atoms:
It is very common that atoms of different elements combine to form compound through covalent bond. For example, H₂, O₂, N₂, F₂ e.t.c.
(i) Ions:
Other than covalent compounds we have ionic compounds. Ionic compounds are made up of ions. These ions forming the ionic compounds can be monatomic like Na⁺, Br⁻, Mg²⁺, Al³⁺, N⁻³ or they can be polyatomic like CO₃²⁻, SO₄²⁻, NH₄⁺, PO₄³⁻ e.t.c.
(iii) Polyatomic Ions:
In polyatomic ions we find a charge on a molecule which contains two or more atoms bonded covalently. Hence, in given compound we have a long chain of molecule containing a negative charge neutralized by opposite +ve charged sodium ion. Hence, Sodium oleate is a polyatomic ionic compound.
Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]
Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess.
Follow the steps as
1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.
n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-
Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess
n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.
Total volume= V acid + V base= 28 ml + 60 ml = 98 ml
Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M
The answer is 0.009 M.
Answer:
Explanation:
Your strategy here will be to
use the chemical formula of carbon dioxide to find the number of molecules of
CO
2
that would contain that many atoms of oxygen
use Avogadro's constant to convert the number of molecules to moles of carbon dioxide
use the molar mass of carbon dioxide to convert the moles to grams
So, you know that one molecule of carbon dioxide contains
one atom of carbon,
1
×
C
two atoms of oxygen,
2
×
O
This means that the given number of atoms of oxygen would correspond to
4.8
⋅
10
22
atoms O
⋅
1 molecule CO
2
2
atoms O
=
2.4
⋅
10
22
molecules CO
2
Now, one mole of any molecular substance contains exactly
6.022
⋅
10
22
molecules of that substance -- this is known as Avogadro's constant.
In your case, the sample of carbon dioxide molecules contains
2.4
⋅
10
22
molecules CO
2
⋅
1 mole CO
2
6.022
⋅
10
23
molecules CO
2
=
0.03985 moles CO
2
Finally, carbon dioxide has a molar mass of
44.01 g mol
−
1
, which means that your sample will have a mass of
0.03985
moles CO
2
⋅
44.01 g
1
mole CO
2
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
a
a
1.8 g
a
a
∣
∣
−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the number of atoms of oxygen present in the sample.
