Correct Question: what is the oxidizing agent in the reaction.
2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Answer: MnO4-is the oxidizing agent
Explanation:
In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Oxidizing agent oxidizes other molecules while the themselves get reduced.
oxidizing agents give away Oxygen to other compounds.
MnO4-is the oxidizing agent because
On the reactants side
Oxidation number of Mn in 2MnO4- is +7
Oxidation number of Cl- is -1
On the products side
Oxidation number of Mn is +2
While oxidation number of Cl is zero
Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2
Answer:
The solution is basic.
Explanation:
We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):
- If no. of millimoles of acid > that of base; the solution is acidic.
- If no. of millimoles of acid = that of base; the solution is neutral.
- If no. of millimoles of acid < that of base; the solution is basic.
- We need to calculate the no. of millimoles of acid and base:
no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.
no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.
<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>
<em>So, the solution is: basic.</em>
Molar mass :
Cl₂ = 71.0 g/mol Na = 23.0 g/mol
<span>2 Na + Cl</span>₂<span> = 2 NaCl
</span>
2 x 23 g Na -------> 71.0 g Cl₂
96.6 g Na ----------> ?
Mass Cl₂ = ( 96.6 x 71.0 ) / ( 2 x 23 )
Mass Cl₂ = 6858.6 / 46
= 149.1 g of Cl₂
hope this helps!
Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol
Molarity: mol/L
First, find mol.
62.5 g x 1mole ÷ 40.31 g = 1.55 mol
then divide mol and the given liters
1.55mol ÷ 1.50 L= 1.03 M
11. D
12.C
13. For 13 i would say B but not so sure
14. A
