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ra1l [238]
3 years ago
12

What is the number of electrons in an electrically neutral atom equivalent to?

Chemistry
2 answers:
dybincka [34]3 years ago
5 0

Answer:

Option B = atomic number

Explanation:

Electrically neutral atom:

An electrically neutral atom have equal number of protons and neutrons. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons.  The number of protons or number of electrons are the atomic number of an atom while the number of protons and number of neutrons are the mass number of an atom. Every atom consist of nucleus or a positive center. The protons and neutrons are present with in the nucleus while electrons are present out side the nucleus. All these three subatomic particles construct an atom.

Example:

Consider the example of carbon. It consist of six protons and six electrons. So, its atomic number is six. While the six neutrons are also present in nucleus with six protons that’s why its mas number is 12.

Electron:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e⁻

Mass= 9.10938356×10⁻³¹ Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

Neutron and proton:

While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.

Symbol of proton= P⁺  

Symbol of neutron= n⁰  

Mass of proton=1.672623×10⁻²⁷Kg

Mass of neutron=1.674929×10⁻²⁷ Kg

Bad White [126]3 years ago
4 0

Answer: atomic number

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Why is it impossible to ever prove that a hypothesis is true?
maria [59]

Answer: A

Explanation:

Hypothesis have to be based on previous knowledge and have to be able to be tested but the problem is most hypothesis can’t be explained so it’s impossible to prove a hypothesis correct because science uses inductive reasoning

4 0
4 years ago
A chemist weighed out 19.9 g of aluminum. Calculate the number of moles of aluminum she weighed out.
vodomira [7]
Aluminum is 26.982 grams per mole, so 26.982/19.9 will give you the moles 1.3558794
3 0
4 years ago
When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t
Harlamova29_29 [7]

Answer: Molarity of MgCl_2 in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{\text{no of moles}}{\text{Volume in L}}

{\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96

Now put all the given values in the formula of molarity, we get

Molarity=\frac{1.96}{1.00L}

Molarity=1.96mol/L

Thus molarity of MgCl_2 in the original sample was 1.96M

4 0
3 years ago
4
Stolb23 [73]

Answer:

Um english please I don’t know Jewish

Explanation:

6 0
3 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
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