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Korolek [52]
3 years ago
7

A mole equals 6.02 x 10^23 . Answer these questions below.

Chemistry
1 answer:
siniylev [52]3 years ago
7 0

Answer:

1. 1.25 mol ants x 6.02*10^23 ants/1 mol ants = 7.53*10^23 ants

2. 4.92*10^26 pencils x 1 mol pencils/6.02*10^23 pencils = 817 mol pencils

3. 0.26 mol molecules x 6.02*10^23 molecules/1 mol molecules = 1.6*10^23 molecules

4. 3.46*10^19 molecules x 1 mol molecules/6.02*10^23 molecules = 5.75*10^-5 mol molecules

5. 5.3*10^20 atoms x 1 mol atoms/6.02*10^23 atoms = 8.8 mol atoms

6. 0.11 mol atoms x 6.02*10^23 atoms/1 mol atoms = 6.6*10^22 atoms

I would suggest looking into "dimensional analysis" for help with this type of material. Dimensional analysis will stick with you all throughout chemistry, so picking it up will be extremely beneficial.

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What is the maximum number of electrons that can be contained in the first, second, third, and fourth energy levels, respectivel
goldenfox [79]
We can find it by 2n^2 formula
So
First shell is maximum of 2
Second shell is of 8
Third shell is of 18
Fourth shell is of 32
8 0
4 years ago
If NaCl has a mass of 3.2g, what is the volume of chlorine gas at STP?
Shalnov [3]

Hey there!

Molar mass NaCl = 58.44 g/mol

58.44 g ----------------- 22.4 ( at STP )

3.2 g -------------------- Volume ??

Volume = ( 3.2 x 22.4 )  / 58.44

Volume = 71.68 / 58.44

Volume = 1.226 L

Hope this helps!

5 0
3 years ago
Read 2 more answers
Tris base has a molecular weight of 121 g/mol. How many grams of tris base would you need to make 250ml of a 200mM SOLUTION
STatiana [176]

Answer:

6.05 g

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

From the question ,

M = 200mM

Since,

1 mM = 10⁻³ M

M = 200 * 10⁻³ M

V = 250 mL

Since,

1 mL = 10⁻³ L

V = 250 * 10⁻³ L

The moles can be calculated , by using the above relation,

M = n / V  

Putting the respective values ,

200 * 10⁻³ M = n / 250 * 10⁻³ L

n = 0.05 mol

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

m = 121 g/mol

n = 0.05 mol ( calculated above )

The mass of tri base can be calculated by using the above equation ,

n = w / m  

Putting the respective values ,

0.05 mol = w / 121 g/mol

w = 0.05 mol * 121 g/mol

w = 6.05 g

3 0
4 years ago
Read 2 more answers
Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an
SVETLANKA909090 [29]

Answer : The standard cell potential of the reaction is, -1.46 V

Explanation :

The given balanced cell reaction is,  

3Pb^{2+}(aq)+2Cr(s)\rightarrow 2Cr^{3+}(aq)+3Pb(s)

Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

The standard values of cell potentials are:

Standard reduction potential of lead E^0_{[Pb^{2+}/Pb]}=-0.13V

Standard reduction potential of chromium E^0_{[Cr^{3+}/Cr]}=1.33V

Now we have to calculate the standard cell potential for the following reaction.

E^0=E^0_{cathode}-E^0_{anode}

E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}

E^0=(-0.13V)-1.33V=-1.46V

Therefore, the standard cell potential of the reaction is, -1.46 V

6 0
4 years ago
If you have 3 moles of calcium carbonate, how many grams of calcium bicarbonate are formed?
Flura [38]

Answer:

hhjcioz xlioyudiyyxyisrupautwtritu regards Roy

7 0
3 years ago
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