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3 years ago

Describe the process of evaporation.

1 answer:

5 0

Hopefully this helps, this is basically simplified but you could ask me for a more detailed explanation. But evaporation works by water on the ground(puddles, lakes, etc) turn into a vapor, which means the water tunrs into gas, which is invisible. Then that water goes up and is stored in the clouds which would then become water and fall into a puddle and the cycle repeats! Hope this Helped!

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A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

What type of plate boundary would form where there is a mountain range present, known earthquake activity, but no trenches prese

Natasha_Volkova [10]

Answer:

continental-continental convergent boundary

Explanation:

When two continental plates converge, they smash together and create mountains. The amazing Himalaya Mountains are the result of this type of convergent plate boundary. The Appalachian Mountains resulted from ancient convergence when Pangaea came together.

5 0

3 years ago

Why is cl2 a subcript

vazorg [7]

Answer:

In Cl $_2$ , the 2 is a subscript because it indicates there are 2 of the same elements. The Lewis structure would display it as Cl-Cl.

On the other hand, a superscript would indicate a specific charge.

All subscripts show the amount of the specific element there is.

An example would be O $_2\\$ or N $_2$ , they both show that there are 2 of the same elements.

If the subscript is outside a parenthesis such as $Ca_3(PO_4)_2$ it indicates there are 2 $PO_4$ molecules.

5 0

3 years ago

Why edible oils shows rancidity when stored for a long time

iren2701 [21]

When edible oils are idle and stored for a long amount of time, they undergo oxidation due to the exposure to oxygen. This oxidation causes rancidity in oils.

4 0

3 years ago

If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theor

Korvikt [17]

Answer:

F2 is the limiting reactant

27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

2Na+ F2 ---> 2NaF

To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

=0.658 moles NaF

16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)

=0.705 moles NaF

Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.

8 0

3 years ago