All categories

3 years ago

How many moles of lithium are in 83 grams of lithium?

Chemistry

2 answers:

german3 years ago

8 0

1 moles lithium or 6.941 grams

Paha777 [63]3 years ago

8 0

You would do 83grams of Li x 1mol of Li/molar mass of lithium (6.94 g/mol)

It would equal 11.96 moles of lithium

You might be interested in

Which types of asexual reproduction could be seen

natita [175]

Answer:

Bacteria

Explanation:

On a piece of raw, fresh hamburger from the butcher's, bacteria are rapidly multiplying through asexual reproduction.

5 0

3 years ago

Caffeine (C8H10N4O2) is a stimulant found in coffees and teas. When dissolved in water, it can accept a proton from a water mole

Olenka [21]

Answer:

See figure 1

Explanation:

If we want to find the acid and the Brønsted-Lowry base, we must remember the definition for each of these molecules:

-) Acid: hydrogen donor

-) Base: hydrogen acceptor

In the caffeine structure, we have several atoms of nitrogen. These nitrogen atoms have the ability to accept hydronium ions ( $H^+$ ). Therefore the caffeine molecule will be the base since it can accept

If caffeine is the base, the water must be the acid. So, the water in this reaction donated a hydronium ion.

Thus, caffeine is the base and water the acid. (See figure 1)

3 0

3 years ago

Salts are which part of an aqueous solution?

masha68 [24]

Explanation:

Salts are the solutes in an aqueous solution. An aqueous solution is solution whose solvent water.

To form a solution,a solute must be dissolved in a solvent.
For aqueous solutions, the solvent which is the dissolving medium is made up of water.
The solute is the substance that is dissolved in it.
Salts for example can be a solute in an aqueous solution.
A salt is generally an ionic compound consisting of positive ions such as metallic, ammonium ans complex ions and negative ions such as acid radicals and complex ions.

Learn more:

Aqueous solution brainly.com/question/8426727

#learnwithBrainly

3 0

3 years ago

Which of the following will occur soon after a population of prey increases slightly in size?

Lady bird [3.3K]

Your answer is D predator population will increase greatly. That's because if there's more prey there will be more food and its good for the predators to produce more because they don't got to worry about food. Speaking how they preys population grew.

So your answer is D.

4 0

3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

7 0

2 years ago