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mr_godi [17]
3 years ago
6

A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next fe

w minutes? Molecules in both the metal and the surrounding air will start moving at lower speeds. Molecules in both the metal and the surrounding air will start moving at higher speeds. The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up. The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down. Mark this and return
Chemistry
1 answer:
spayn [35]3 years ago
5 0
Cyndi hang Geld Robin
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In solids the interatomic or intermolecular space is least . It is most pronounced in gases . That is why inter molecular or interatomic attraction is least in gases . That is why gas flows .

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When you decrease the volume of a gas, what happens to the temperature?
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the temperature is decreased

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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.

3 0
3 years ago
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