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Drupady [299]
3 years ago
11

What is the reduction half-reaction for the following unbalanced redox equation?

Chemistry
2 answers:
bazaltina [42]3 years ago
5 0
First find the oxidation states of the various atoms: 
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>

<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Reptile [31]3 years ago
4 0
C. N2 → NH4+ would be the correct option
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Mixture of same boiling point cannot be separated by using distillation why?<br>​
Bond [772]

Answer:

both the substances will evaporate

Explanation:

3 0
3 years ago
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I know how to solve it with D=M/V and M1V1 however the answer isn’t correct. Help me please
lara31 [8.8K]

Answer:

23.28 g of O2.

Explanation:

We'll begin by calculating the mass of hexane. This can obtain as follow:

Volume of hexane = 10 mL

Density of hexane = 0.66 g/mL

Mass of hexane =?

Density = mass /volume

0.66 = mass of hexane /10

Cross multiply

Mass of hexane = 0.66 x 10

Mass of hexane = 6.6 g

Next, we shall write the balanced equation for the reaction. This is given below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:

Molar mass of C6H14 = (12.01x6) + (1.008 x 14)

= 72.06 + 14.112

= 86.172 g/mol

Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608 g

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.

Therefore, 23.28 g of O2 is needed for the reaction.

8 0
3 years ago
Which of the following are sources of revenue for media companies? A. Direct sales to producers B.advertising and subscriptions
Fed [463]

Answer:

c

Explanation:

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4 years ago
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What is the volume of 40g of sugar given it’s listed density
lora16 [44]

Answer:

25.157 cm³

Explanation:

Data Given:

Mass of Sugar (m) = 40g

Density of sugar given in literature = 1.59 g/cm³

Volume of Sugar = ?

The formula will be used is

                              d = m/v ........................................... (1)

where

D is density

m is the mass

v is the volume

So

Rearrange the Equation (1)

                              d x v = m

                               v = m/ d         ................................................ (2)

put the given values in Equation  (2)

                       v = 40g / 1.59 g/cm³

                       v = 25.157 cm³

volume of 40 g of sugar = 25.157 cm³

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