What are some geographic features that could be found in the hydrosphere?

Chemistry

1 answer:

nadezda [96]3 years ago

6 0

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

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Differentiate between mass, volume, and density. Explain why only changing the shape of an object would have no effect on its de

Anni [7]

Mass is how much matter is in an object. Volume is how much space an object occupies. Density is the amount of mass in a volume.

Have a nice day, brainliest would be fantastic

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3 years ago

How many grams of aluminum are in 0.58 moles of Al2O3.2H:0?

Pani-rosa [81]

Answer:

80.04 gram

Explanation:

mass = moles × relative formula mass

= 0.58×138

= 80.04

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4 years ago

Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to

Mama L [17]

We have the given reaction as;

$HCl_{(aq)} + NaOH_{(aq)}$ ----> $NaCl_{(aq)} + H_{2}O_{(l)}$ 

Answer A) The pH will be 12.36,

We have to convert the concentrations of HCl and NaOH into moles,

So we have, n(HCl) = (0.0290 L) X (0.290 mol/L) = 8.41X $10^{-3}$ moles 

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X $10^{-2}$ moles

Now, it seems NaOH is in excess, so amount remaining will be;

1.13 X $10^{-2}$ - 8.41 X $10^{-3}$ = 2.89 X $10^{-3}$ moles

Now, the total volume will become as = 0.0390 + 0.0290 = 0.068 L 

So, the concentration of [ $OH^{-}$ ] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X $10^{-2}$ M

pOH = - log [ $OH^{-}$ ] = -log (4.25× $10^{-2}$ ) = 1.37

Hence, pH = 14 - pOH = 14 - 1.37 = 12.6

So the pH of the solution will be 12.6 which is basic in nature. 

Answer B) The pH will be 1.68

Now, for the given concentration we need to find moles for HCl and NaOH also;

n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X $10^{-3}$ mol 

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X $10^{-3}$ mol 

here we can see, HCl is in excess amount so the remaining will be;

8.41X $10^{-3}$ - 7.41 X $10^{-3}$ = 1.0 X $10^{-3}$ mol

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L

So the concentration of [HCl] = 1.0 X $10^{-3}$ mol / 0.0480 L = 2.08 X $10^{-2}$ M

Which is = [H⁺] 

So, the pH = - log [ $H^{+}$ ] = -log(2.08X $10^{-2}$ ) = 1.68