When an element exist in nature by itself, it must have a charge of

In a classroom, which comparison would a teacher most likely use for describing a mole?

a_sh-v [17]

Answer:

In a classroom, the comparison that a teacher would most likely use for describing a mole is: jar of jellybeans and a gram of salt crystals.

Explanation:

4 0

3 years ago

What is the pOH of a solution with [OH-] = 1.4 x 10-13?

lukranit [14]

Answer:

C. 12.85

Explanation:

just did

4 0

3 years ago

A mixture 21.7 g NaCl 3.74 g kcl and 9.76 g licl how many moles of nacl are in this mixture

slava [35]

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of NaCl = 21.7 g
Molar mass of NaCl = 58.4 g/mol
Hence, moles of NaCl = 21.7 g / 58.4 g/mol = 0.372 mol

Hence moles of NaCl in the mixture is 0.372 mol.

Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.

mole % = (moles of desired substance / Total moles of the mixture) x 100%

Hence,
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%

Total moles of mixture = moles of NaCl + KCl + LiCl

Mass of KCl = 3.74 g
Molar mass of NaCl = 74.6 g/mol
Hence, moles of NaCl = 3.74 g / 74.6 g/mol = 0.050 mol

Mass of NaCl = 9.76 g
Molar mass of NaCl = 42.4 g/mol
Hence, moles of NaCl = 9.76 g / 42.4 g/mol = 0.230 mol

Total moles = 0.372 mol + 0.050 mol + 0.230 mol = 0.652 mol

mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
= (0.372 mol / 0.652 mol) x 100%
= 57.06%

Hence, mixture has 57.06% of NaCl as the mole percentage.

8 0

3 years ago

How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2

jeyben [28]

Answer: The amount of barium sulfate produced in the given reaction is 0.667 grams.

Explanation:

To calculate the number of moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol$

For the given chemical reaction:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = $\frac{1}{1}\times 0.00286mol=0.00286mol$ of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

$0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g$

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

4 0

4 years ago

How does the energy of an electron change when the electron moves closer to the nucleus?

tankabanditka [31]

It increases. B.

Close to the nucleus are high energy levels, because of the pull of the nucleus. But farther away from the nucleus, the electrons there are of low energy level.

5 0

4 years ago