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3 years ago

PLEASE HELP!

Chemistry

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

The Axis Powers took these aggressive actions prior to the outbreak of World War II which include the invasion of Ethiopia, Manchuria, and the Rhineland.

Explanation:

Axis Powers: It was the coalition of Germany, Italy and Japan against the <em>Allied Powers</em> (the United States, the Soviet Union, Great Britain, and China) in World War II.

In 1930, the Japanese took control of railway track in Manchuria (a region in China) and in 1931, the Japan attacked the Chinese army claiming that the Chinese soldiers sabotaged the railway.
On 3 October 1935, Italy invaded the Ethiopia (former name: Abyssinia) in the leadership of <em>Mussolini </em>to expand its territories.
Germany in the leadership of Adolf Hitler occupied the Rhineland in 1937 by making excuse that the area is hostile and dangerous for them.

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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.

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babymother [125]

Answer:

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densk [106]

By looking at the stochiometry, we can conclude that the ratio is 1 :1

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