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3 years ago

Which is earths largest source of drinkable water

Chemistry

1 answer:

Hitman42 [59]3 years ago

8 0

Answer:

The Ocean

Explanation:

The Ocean is what makes up most of the water on Earth which makes it the largest source of water you can find. As long as you make the Ocean/Sea water drinkable (by treating/helping to remove dirt, salt and other harmful components) then you should be fine.

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Determine the number of atoms in 41.0 grams of calcium, Ca. (The mass of one mole of calcium is 40.08 g.)

Dovator [93]

Hi there!

Let me know if you have questions about my answer:

6.15 × 10²⁴

Explanation:

To find the number of atoms, divide the mass of calcium (m) by its molar mass (MM). This will give you the number of moles of calcium (n).

$n = m/MM$ Start with a formula

$=\frac{41.0g}{40.08g/mol}$ Substitute values from the question

$= \frac{41.0g*1mol}{40.08g}$ Simplify

$= \frac{41.0mol}{40.08}$ Cancel out the units "g"

$= 1.022...mol$ Number of moles of calcium, with 4 significant figures

Then, multiply the number of moles of calcium (n) by Avogadro's number ( $N_{A}$ ) to find the number of calcium atoms.

$Number.of.Ca=n*N_{A}$ Start with a formula

$= 1.022mol*\frac{6.022*10^{23}Ca}{1mol}$ Substitute values from the question

$= \frac{1.022mol*6.022*10^{23}Ca}{1mol}$ Simplify

$= 1.022*(6.022*10^{23}Ca)$ Cancel out the units "mol"

$= 6.154*10^{24} Ca$ Multiply, leave one extra digit to help round. "4" tells you to round down.

$= 6.15*10^{24} Ca$ Final answer to 3 significant figures*

*When multiplying and dividing numbers, your final answer will have the same number of significant figures as the number with the least number of significant figures in the question.

40.08 has 4 significant figures. 41.0 has 3 significant figures. So, you want 3 significant figures in your final answer.

I hope this helps!

Learn more about Avogadro's number here:

brainly.com/question/1445383

Check out another answer that uses significant figures:

brainly.com/question/3721164

7 0

3 years ago

¿Cuál es la cantidad de electrones (e-) de Níquel si tiene una masa atómica de 58.6 y un número

daser333 [38]

Answer:

Hi do we translate a this

Explanation:

4 0

3 years ago

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0

3 years ago

A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was dete

Katarina [22]

We will use boiling point formula:

ΔT = i Kb m

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass Kg

0.27 = moles / 0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol

5 0

3 years ago

I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti

Tanya [424]

<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>

4 0

3 years ago