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statuscvo [17]
4 years ago
8

I don’t understand this, can you help me?

Chemistry
1 answer:
vlada-n [284]4 years ago
3 0
An atom becomes stable if it has a full and complete ring. 2 of the choices are wrong off the start because you can't gain electrons and have a net positive charge, nor can you lose electrons and have a net negative charge. That leaves us with Option A or D. As I said earlier, an atom becomes stable if it has a full and complete ring. Obviously, if you lose 3 electrons from the last ring, the atom would be stable (because the 2nd ring is now full), and it also would have a net positive charge because it lost electrons.
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For it to be the same element it must contain the same number of protons
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An example of a solid solution is copper true or false
garik1379 [7]
This is false, because copper is an element.
To be solution, it has to have at least 2 different substances. Solid solution of metals is often called alloy. 
4 0
3 years ago
What frequency does television channel 7 broadcast at?
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3 years ago
Read 2 more answers
For the reaction: 3 KClO3 to 3 KCL + 4O2, How many miles of potassium chloride are produced from 9.3 grams of potassium chloride
Valentin [98]

5.7 g of KCl is produced.

<u>Explanation:</u>

First we have to write the balanced equation as,

2 K C l O_{3} \rightarrow 2 K C l+3 O_{2} \uparrow

Then we have to find the moles of potassium chlorate as,

Moles of $K C l O_{3}=9.3 g$ of $K C l O_{3} \times \frac{1 \text { mol of } K C l O_{3}}{122.55 g K C l O_{3}}=0.076$ moles

Then we have to find the moles of KCl as,

Moles of $\mathrm{KCl}=0.076$ moles of $\mathrm{KClO}_{3} \times \frac{2 \text { mol of } \mathrm{KCl}}{2 \text { mol } \mathrm{KClo}_{3}}=0.076$ moles

From the moles of KCl, we have to find the mass using its molar mass by multiplying them as,

\text { Mass of } \mathrm{KCl}=0.076 \text { mol } \times 74.55 \frac{g}{\mathrm{mol}}=5.7 \mathrm{g} \text { of } \mathrm{KCl}

3 0
3 years ago
5 drops of 0.15 M Ki added to<br>40 drops of Na2S2O3<br>What is the final concentration of ki?​
rjkz [21]

Answer:

\boxed{\text{0.017 mol/L}}

Explanation:

Na₂S₂O₃ solution does not react with KI (it reacts with I₂), so it is simply diluting the KI, and we can use the dilution formula.

c_{1}V_{1} = c_{2}V_{2}

Data:

c₁ = 0.15 mol·L⁻¹; V₁ = 5 drops

V(Na₂S₂O₃) = 40 drops

Calculations:

(a) Calculate the total volume

V₂ = 5 + 40 = 45 drops

(b) Calculate the concentration

0.15 × 5 = c₂ × 45

0.75 = 45c₂

c_{2} = \dfrac{0.75 }{45} = \boxed{\textbf{0.017 mol/L}}

3 0
3 years ago
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