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AlekseyPX
3 years ago
12

A beaker contains 500 ml of 2.40M KNO3 the beaker is left uncovered so that all of the water evaporates. What mass of KNO3 cryst

als will remain in the beaker when the beaker is dry.
Chemistry
1 answer:
telo118 [61]3 years ago
6 0

Answer:

242.4 g

Explanation:

RAM of KNO3=39+14+(16×3)=101

Mass=morality×RAM

101*2.4=242.4

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What is the most likely way a chemist would be involved in the production of<br> a computer chip?
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3 years ago
Which ions remain in solution after pbi2 precipitation is complete? express your answers as ions separated by a comma?
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Answer;

K+ and NO3- ions

Explanation;

The main ions remaining are K+ and NO3- ions after pbi2 precipitation is complete.

However; There will always be tiny amounts of Pb2+ and I- ions, but most of them are in the solid precipitate.

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3 years ago
Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) → 2Al2O3 (s) In
jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol

For oxygen gas:

\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol

The chemical equation for the reaction of titanium and chlorine gas follows:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = \frac{3}{4}\times 0.092=0.069mol of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of Al_2O_3

So, 0.092 moles of aluminium will produce = \frac{2}{4}\times 0.092=0.046moles of Al_2O_3

Now, calculating the mass of aluminium oxide:

\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield  = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%

Hence, the percent yield of the reaction is 74 %

6 0
3 years ago
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