Can someone plz help me in this thanks for number 1. Only

1.For the first ten reactions, write a balanced reaction for each tube in which a reaction occurred. If there was no reaction, w

slamgirl [31]

The equations of reaction occurring in the tubes are as follows:

2 MnO₄⁻ + 6 Br⁻ + 8 H⁺ → 2 MnO₂ + 3 Br₂ + 4 H₂O
2 MnO₄⁻ + 6 I⁻ + 8 H⁺ → 2 MnO₂ + 3 I₂ + 4 H₂O
No reaction
2 Fe³⁺ + 2 I⁻ → 2 Fe²⁺ + I₂

<h3>What are the reactions occurring in the tubes?</h3>

The reactions occurring in the tubes are redox reactions.

Based on the table the equations of reaction are as follows:

2 MnO₄⁻ + 6 Br⁻ + 8 H⁺ → 2 MnO₂ + 3 Br₂ + 4 H₂O
2 MnO₄⁻ + 6 I⁻ + 8 H⁺ → 2 MnO₂ + 3 I₂ + 4 H₂O
No reaction
2 Fe³⁺ + 2 I⁻ → 2 Fe²⁺ + I₂

In conclusion, redox reaction are reactions in which electrons are transferred.

Learn more about redox reactions at: brainly.com/question/26750732

#SPJ1

6 0

1 year ago

Which element forms compounds which are involved in smog and acid rain?

Tcecarenko [31]

Answer:

c. nitrogen

Explanation:

5 0

2 years ago

Read 2 more answers

Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains

kiruha [24]

Answer:

Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)

If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C

Using dc = (1/4πεo)qQ/Eα we have

dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm

Note: 1meter = 10^15fentometer

Explanation:

This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.

7 0

3 years ago

According to Newton's third law, forces come in pairs. There is_____and an equal and opposite_____.

hichkok12 [17]

According to Newton's third law, forces come in pairs. There is action-reaction force and an equal (in size - action force) and opposite in direction (reaction force).

<u>Explanation:</u>

On the off chance, when item A applies a force on item B, at that point item B must apply a force of equivalent size and inverse bearing back on object A. This Newton's third law speaks to a specific balance in nature: forces consistently happen two by two.

And, one body can't apply a force on another without encountering a force itself. Now and again, allude to this law freely as activity response, where the force applied is the activity and the force experienced as a result is the response.

4 0

2 years ago

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add p

Butoxors [25]

Answer:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Explanation:

Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

$BiO_3^-\rightarrow Bi^{3+}$

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

$(Bi^{5+}O_3)^-\rightarrow Bi^{3+}$

Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

$6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O$

Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

$6OH^-+6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Then, we accommodate the waters to obtain:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Best regards!

7 0

3 years ago