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NeTakaya
2 years ago
12

Choose all the answers that apply. Lakes are recharged by _____. 1.evaporation 2.precipitation 3.rivers 4.transpiration 5.ground

water seepage 6.melting snow and ice
Chemistry
2 answers:
sweet-ann [11.9K]2 years ago
6 0
2. Precipitation, because when it rains the water that was taken when it was evaporated is replaced and it fills the lake back up
julsineya [31]2 years ago
6 0
THE ANSWER IS PRECIPITATION. PRECIPITATION IS RAIN. EVERY TIME IT RAINS, EVERY BODY OF WATER IS INVOLVED IN THIS PROCESS. THE OCEAN, LAKES, PONDS, ETC
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How many liters of oxygen gas can be produced at stp from the decomposition of 0.250 l of 3.00 m h2o2 in the reaction according
Wewaii [24]

the balanced chemical equation for the decomposition of H₂O₂ is as follows

2H₂O₂ ---> 2H₂O + O₂

stoichiometry of H₂O₂ to O₂ is 2:1

the number of moles of H₂O₂ decomposed is - 0.250 L x 3.00 mol/L = 0.75 mol

according to stoichiometry the number of O₂ moles is half the number of H₂O₂ moles decomposed

number of moles of O₂ - 0.75 mol / 2 = 0.375 mol

apply the ideal gas law equation to find the volume

PV = nRT

where P - standard pressure - 10⁵ Pa

V - volume

n - number of moles 0.375 mol

R - universal gas constant - 8.314 Jmol⁻¹K⁻¹

T - standard temperature - 273 K

substituting the values in the equation

10⁵ Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K

V = 8.5 L

volume of O₂ gas is 8.5 L

6 0
2 years ago
A. The measured pH of a 0.100 M HCl solution at 25 degrees Celsius is 1.092. From this information, calculate the activity coeff
ra1l [238]

Answer:

activity coefficient \mathbf{\gamma =0.809}

activity coefficient \mathbf{\gamma = 0.791}

The change in pH in part A = 0.092

The change in pH in part B =  0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] = 10^{-1.092}

[a] = 0.0809 M

Recall that [a] = \gamma × C

where;

\gamma  = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

\gamma = \dfrac{[a]}{C}

\gamma = \dfrac{0.0809 \ M}{0.100  \ M}

\mathbf{\gamma =0.809}

B.

The pH of a solution of HCl and KCl = 2.102

[a] = 10^{-2.102}

[a] = 0.00791 M

activity coefficient \gamma = \dfrac{0.00791 \ M}{0.01 \ M}

\mathbf{\gamma = 0.791}

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102

3 0
2 years ago
When you apply heat energy to a substance, where does the energy go? Think about the law of conservation of energy.
MAVERICK [17]

The energy is transformed into kinetic energy which makes the substance to move. The law of conservation of energy which is the first law of thermodynamics states that in a closed system energy can neither be created nor destroyed but can change from one form to another

7 0
2 years ago
A 0.0035 M aqueous solution of a particular compound has pH = 2.46. The compound is (A) a weak base (B) a weak acid (C) a strong
slava [35]

Answer:

(a) A strong acid

Explanation:

We have given the pH of the solution is 2.46

pH=2.46  

So the concentration of H^+=10^{-pH}=10^{-2.46}=0.00346

solution having H+ concentration more than H^+=10^{-7} is acidic

Since in the given solution, H+ concentration is 0.00346 M which is more than 10^{-7}[/tex] so this is an acidic solution

Note-The concentration of H^+ decide the behavior of the solution that is, it is acidic or basic

7 0
2 years ago
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
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