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Pepsi [2]
3 years ago
5

Can someone answer 1 For me please?

Chemistry
1 answer:
VikaD [51]3 years ago
6 0

Answer:

The answer is c.

Explanation:

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Please Help 8-9!!!!!!!
seropon [69]

The labeled diagram is given in the image attached.

As it can be seen from the image that freezing is when energy is removed from the system at 0 ⁰ while melting is when energy is added at 0⁰.

Also when energy is added at 100⁰C, it causes boiling while when it is removed at 100⁰C, it causes condensation.


Melting point of water is 0⁰C while boiling point is 100⁰C

3 0
2 years ago
(Upvote) Two atoms that are isotopes of one another must have the same number of what? A. Electrons B. All Particles C. Protons
UNO [17]
Protons, the number of protons cannot change or the element will change as well.
4 0
3 years ago
Nickel + oxygen = nickel oxide. What is the balanced redox reaction?
uranmaximum [27]

Answer:

This is an oxidation-reduction (redox) reaction:

2 Ni0 - 4 e- → 2 NiII

(oxidation)

2 O0 + 4 e- → 2 O-II

(reduction)

Ni is a reducing agent, O2 is an oxidizing agent.

7 0
2 years ago
What is nuclear fusion what role does it play in the sun
GalinKa [24]

Answer:

Nuclear fusion is when two atomic nuclei combine and form one nucleus. Nuclear fusion generates all of the Sun's energy. Inward pulling by the Sun's gravity is counteracted by outward pushing by the Sun's nuclear fusion; this balance keeps the sun from collapsing or exploding.

Explanation:

5 0
2 years ago
Read 2 more answers
Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc
Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

3 0
3 years ago
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