The given question is incomplete. The complete question is :
In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)
1. 
here underlined is 
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
2. 
Here underlined is 
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
3. 
Here underlined is 
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
Answer: 1. Brønsted-Lowry acid
2. Lewis base
3. Brønsted-Lowry base
Explanation:
According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
1. 
As 
is donating a proton , it acts as a bronsted acid.
2.
As contains a lone pair of electron on nitrogen , it can easily donate electrons to 
and act as lewi base.
3.
As 
is accepting a proton , it acts as a bronsted base.
Answer:
1.06 V
Explanation:
The standard reduction potentials are:
Ag^+/Ag E° = 0.7996 V
Ni^2+/Ni E° = -0.257 V
The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are
Ni → Ni^2+ + 2e- E° = 0.257 V
<u>2Ag^+ 2e- → 2Ag </u> <u>E° = 0.7996 V
</u>
Ni + 2Ag^+ → Ni^2+ + 2Ag E° = 1.0566 V
To three significant figures, the standard potential for the cell is 1.06 V
.
Answer:- HBr is limiting reactant.
Solution:- The given balanced equation is:
From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.
The calculations could be shown as:
= 24 mol HBr
From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.
Answer:
One AMU or 1 Atomic Mass Unit
Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
