9.25x10^-8 x6.40x10^3 in scientific equation

What is the oxidizing agent in the reaction 2MnO4-(aq) 10I-(aq) 16H (aq) ---> 2Mn2 (aq) 5I2(aq) 8H2O(l)

faltersainse [42]

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

4 0

3 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

3 years ago

A buffer consists of 0.45 M CH3COOH (acetic acid) and 0.35 M CH3COONa. The Ka of acetic acid is 1.8 x 10-5 . a) Calculate the pH

Zigmanuir [339]

Answer:

A) pH of Buffer solution = 4.59

B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original buffer solution = 4.65

Explanation:

This is the Henderson-Hasselbalch Equation:

$pH = pKa + log\frac{[conjugate base]}{[acid]}$

to calculate the pH of the following Buffer solutions.

8 0

3 years ago

Compare the type of change that occurs when an iron bar rusts and when a substance freezes.

Ivanshal [37]

Explanation:

when an iron bar rust is an example of a chemical change in which a new substance is formed and the change is not easily reversible.for iron to rust moisture and air must be present.while when a substance freezes,it can be easily reversed through melting and no new substance is formed.this change is termed a physical change.

5 0

2 years ago

4 Weathering Assessment

alexira [117]

Answer:

This is an example of chemical change because the Kool aid dissolved and turned the water red

6 0

3 years ago