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nikitadnepr [17]
3 years ago
13

Which of the following accurately pairs the part of an atom to its charge? A. Electron—no charge B. Proton—no charge C. Neutron—

positive charge D. Electron—negative charge
Chemistry
1 answer:
Aneli [31]3 years ago
7 0

Answer:

D: Electron - Negative Charge

Explanation:

Protons have a positive charge, (pro = +)

Neutrons have a negative charge,  (neutral = 0

Electrons have a negative charge.

An element is determent by the number of protons in the atom, the Isotope of an element is determined by the number of neutrons, and the charge of an atom is determined by its electrons, which is equal to the number of protons.

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What ions are formed when naoh is dissolved in water?.
valentina_108 [34]

Answer:

When sodium hydroxide (NaOH) dissolves in water, it separates into positively - charged sodium ions (cations) and negatively - charged hydroxide ions (anions).

Hope this helped.

7 0
2 years ago
Which cannot be chemically broken down into simpler substances?
Anna71 [15]

Answer:

B.

Explanation:

elements

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What does the atomic number represent?
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C. The number of protons in the atoms nucleus.
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Which atom or ion is the largest?<br><br> A. K<br> B. K+<br> C. Ca<br> D. Ca2+<br> E. Li
pishuonlain [190]

Answer:

A. K

Step-by-step explanation:

Remember the trends in the Periodic Table:

  • Atomic radii <em>decrease</em> from left to right across a Period.
  • Atomic radii <em>increase</em> from top to bottom in a Group.
  • Ionic radii of metal cations are <em>smaller</em> than those of their atoms.

Thus, the largest atoms are in the lower left corner of the Periodic Table.

The diagram below shows that K is closest to the lower left, so it is the largest atom. It is also larger than any of the cations.

5 0
3 years 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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