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sasho [114]
2 years ago
15

3. You are a water molecule in the ocean and are getting ready to move through the water cycle. Please describe your journey thr

ough the cycle. Include all of the following processes: Evaporation, Condensation, Precipitation, Transpiration, Runoff, and Infiltration. Your journey should also include any collection areas you pass through or end up in and the forces (sun or gravity) that is acting on you. Don’t forget to say which state changes you go through (solid, liquid, gas). for 50 points
Chemistry
1 answer:
Usimov [2.4K]2 years ago
8 0

Answer:

Water moves from the ground or oceans into the atmosphere through a process called evaporation. It's a process that happens on a molecular level when the molecules of water are really energized and rise into the air. Now you've got water in the air and water on land. Organisms all over the Earth need water to survive.

Explanation:

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A higher than normal high tide is called
alina1380 [7]
It is called a spring tide
8 0
2 years ago
The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this
azamat

Answer : The value of K_{sp} for BaCO_3 is 19.36\times 10^{-10}mole^2/L^2.

Solution : Given,

Solubility of BaCO_3 in water = 4.4\times 10^{-5}mole/L

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

                            BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3

Initially                   -                   0        0

At equilibrium       -                   s         s

The Solubility product will be equal to,

K_{sp}=[Ba^{2+}][CO^{2-}_3]

K_{sp}=s\times s=s^2

[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L

Now put all the given values in this expression, we get the value of solubility constant.

K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2

Therefore, the value of K_{sp} for BaCO_3 is 19.36\times 10^{-10}mole^2/L^2.

3 0
3 years ago
A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2
Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

<u>Step 1: </u>Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

<u>Step 2</u>: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

<u>Step 3:</u> Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

<u>Step 4: </u>Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

<u>Step 5:</u> Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

4 0
3 years ago
The cell theory was construted by​
NeTakaya

Answer:

Matthias Jakob Schleiden

Explanation:

The cell was initially discovered by Robert Hooke, but Schleiden developed the theory.

Hope this helps! :)

4 0
3 years ago
Given these reactions:
Liula [17]

Answer:

-304kJ

Explanation:

;)

3 0
2 years ago
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