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rewona [7]
2 years ago
8

NaF + HCL=> NACL + HF

Chemistry
1 answer:
Jobisdone [24]2 years ago
4 0

Answer:

NaF + HCl --> NaCl + HF

Explanation:

hope it helps you

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What is a Spontaneous charge
ddd [48]

This means a release of free energy from the system corresponds to a negative change in free energy, but to a positive change for the surroundings.

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3 years ago
What is a product of science
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Answer:

A product in science is a substance that is formed when two or more chemicals react.

Explanation:

When a chemical reaction takes place, a new substance is often created from the atoms or molecules of the original substances. There are often multiple products formed in a reaction.

6 0
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8. Bring the balloon in contact with the wall. What happens to the charges in the wall?
Ne4ueva [31]

When the charged balloon is brought near the wall, it repels some of the negatively charged electrons in that part of the wall. Therefore, that part of the wall is left repelled.

<u>Explanation</u>:  

  • Balloons don't stick to walls. However, if you rub the balloon on an appropriate piece of material such as clothing or a wall, electrons are pulled from the other material to the balloon.
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7 0
3 years ago
Please help,<br> How could you tell a Ca(NO3)2 solution from a Zn(NO3)2 solution?
tamaranim1 [39]

Answer:

you better give me brainliest

Explanation:

Zinc nitrate and calcium nitrate solution can be distinguished by reaction with ammonium hydroxide. Zinc forms a white gelatinous ppt. whereas there is no precipitation of calcium hydroxide even with excess of ammonium hydroxide

3 0
1 year ago
When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
Feliz [49]

Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

8 0
3 years ago
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