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Luden [163]
3 years ago
7

John is outstanding forward on his club select soccer team. He is know for his ball control and speed. After a game he often mak

es a sketch of moves that were successful. The dotted line on his sketch shows the direction the ball was moving before he kicked it.
Chemistry
1 answer:
natima [27]3 years ago
4 0

Answer:

This self evaluation helps them to better his sport.

Explanation:

This is the right way to evaluate the performance of John because this sketch shows him the way he perform in the match. Due to this sketch he can better his performance and remove all the mistakes he had done in the match which results in better performance of John in the next match. This self evaluation helps them to better his sport and got new chances in order to select for a better club or national soccer team.

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Vikki [24]
Fluorine - Seven electrons of it's own. 

Lithium would give up one electron, so there for, fluorine is then left with eight.
3 0
3 years ago
Hey guys!
melamori03 [73]

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7 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
URGENT
antoniya [11.8K]

Answer:

a

Explanation:

8 0
3 years ago
Can someone help me pls
BartSMP [9]

I can help. What is wrong.

7 0
3 years ago
Read 2 more answers
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