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Lunna [17]
3 years ago
5

Given 40 grams of magnesium. How many mole is this

Chemistry
1 answer:
Dahasolnce [82]3 years ago
3 0

To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.

We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'

\frac{40 g Mg}{1} x\frac{1 mole Mg}{24.305 g Mg}

In this, we can see that the unit 'grams' will cancel out to leave us with moles.

In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.

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Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t
Sedbober [7]

Answer:

Approximately 6.30\times 10^{-3}\;\rm mol.

Explanation:

The gallium here is likely to be produced from a \rm NaGaO_2\, (aq) solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral "\mathtt{(III)}" next to \rm Ga.  This numeral indicates that the oxidation state of the gallium in this solution is equal to +3. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of \rm Ga\, (s).

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium \mathtt{(III)} solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s.

Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C.

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}.

It takes three moles of electrons to deposit one mole of gallium atoms \rm Ga\, (s). As a result, \rm 1.89\times 10^{-2}\; mol of electrons would deposit \displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol of gallium atoms \rm Ga\, (s).

8 0
2 years ago
Why are valence electrons important?
Usimov [2.4K]
They are the outer layer of the electron layers.
5 0
3 years ago
Read 2 more answers
Serine has pka1 = 2.21 and pka2 = 9.15. use the henderson-hasselbalch equation to calculate the ratio neutral form/protonated fo
malfutka [58]
Calculate the ratio by using Henderson-Hasselbalch equation:

pH = pKa + log [neutral form] / Protonated form

3.05 = 2.21 + log [neutral form] / [Protonated form]

3.05 - 2.21 = log [neutral form] / [Protonated form]

0.84 = log [neutral form] / [Protonated form]

[neutral form] / [protonated form] = anti log 0.84 = 6.91
8 0
3 years ago
How many grams are in 3.14 x 1015 molecules of CO?
Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15        Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15             Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x                        Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0
2 years ago
A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of
blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of H^+ ion = Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol

Moles of OH^- ion = Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol

HCl+NaOH\rightarrow NaCl+H_2O

For neutralization:

1 mole of H^+ ion will react with 1 mole of OH^- ion

0.01 mol of H^+  ion will react with =\frac{1}{1}\times 0.01mole of OH^- ion

Thus (0.012-0.01)= 0.002 moles of OH^- are left in 20 ml or 0.02 L of solution.

[OH^-]=\frac{0.002}{0.02L}=0.1M

pOH=-log[OH^-]

pOH=-log[0.1]=1

pH+pOH=14

pH=14-1=13

Thus the pH of resulting solution will be 13

7 0
2 years ago
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