All categories

3 years ago

How does most of the water in the water cycle move from lakes and rivers directly back into the atmosphere?

Chemistry

2 answers:

olga2289 [7]3 years ago

8 0

It happens through precipitation when water evaporates from the suns rays and rises into the atmosphere

AnnyKZ [126]3 years ago

7 0

It goes from lakes and as the temp rises, it evaporates into clouds. from there, the clouds get heavy, and it produces rain, and the rain forms rivers and then back into lakes.

You might be interested in

When one mole of a solute is present in 500 cm3 of solution, then the concentration of the solution is:

Schach [20]

Answer:

0.5M is the answer.

Explanation:

1M solution is the solution containing 1mole solute dissolved per litre of solution.

Using unitary method,

1000cc gives 1M.

1cc gives 1/1000M.

500 cc gives 500/1000M=0.5M

7 0

3 years ago

How many moles of ammonia (nh3 will be produced from 4.0 moles of nitrogen (n2?

Lana71 [14]

Standard equation would be N2(g)+3H2(g)==>2NH3(g), so through stoichiometry, (4 mol N2)(2mol NH3/1 mol N2), assuming excess H2, would yield 8 moles of NH3.

3 0

3 years ago

2Fe(OH)3 Fe2O3+3H2O how many grams of Fe2O3 are produced if 10.7 grams of Fe (OH)3 react in this way

sasho [114]

Answer: 7.98 grams of $Fe_2O_3$ are produced if 10.7 grams of $Fe(OH)_3$ are reacted.

Explanation:

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

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

Putting values in equation 1, we get:

$\text{Moles of} Fe(OH)_3=\frac{10.7g}{106.87g/mol}=0.100mol$

The chemical equation for the reaction is

$2Fe(OH)_3\rightarrow Fe_2O_3+3H_2O$

By Stoichiometry of the reaction:

2 moles of $Fe(OH)_3$ produce = 1 mole of $Fe_2O_3$

So, 0.100 moles of $Fe(OH)_3$ produce= $\frac{1}{2}\times 0.100=0.05mol$ of $Fe_2O_3$

Mass of $Fe_2O_3$ = $moles\times {\text{Molar Mass}}=0.05mol\times 159.69g/mol=7.98g$

Hence 7.98 grams of $Fe_2O_3$ are produced if 10.7 grams of $Fe(OH)_3$ are reacted.

3 0

2 years ago

An important natural and metallic resource is ?

ss7ja [257]

The answer is B.) Carbon

3 0

2 years ago

Read 2 more answers

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago