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klemol [59]
1 year ago
12

6. Determine the amount in moles of the following:a. 12.15 g Mgb. 1.50 x 1023 atoms F​

Chemistry
1 answer:
igomit [66]1 year ago
3 0

a. 0.51 moles of Mg

b. 0.25 moles of F

Explanation:

a. To find the number of moles knowing the mass we use the following formula:

number of moles =  mass / molecular weight

number of moles of Mg = 12.15 / 24 = 0.51 moles

b. To find the number of moles knowing the number of atoms we use Avogadro's number to illustrate the following reasoning:

if in        1 mole of F there are 6.022 × 10²³ atoms of F

then in   X moles of F there are 1.5 × 10²³ atoms of F

X = (1 × 1.5 × 10²³) / 6.022 × 10²³

X = 0.25 moles of F

Learn more about:

Avogadro's number

brainly.com/question/13759899

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If you have ever peeled the label off a glass jar, you may have noticed that the glue does not easily wash off with water. Howev
shepuryov [24]

Answer:

It can be removed by acidic chemicals

Explanation:

8 0
2 years ago
5. You have 100 kg of gas of the following composition: CH4 - 30% H2 - 10% N2 - 60% What is the average molecular weight of this
Lorico [155]

Answer:

21.8 g/mol

Explanation:

Molecular weight of CH4 = 16g/mol

H2 = 2g/mol

N2 = 28g/mol

(16*30 + 2*10 + 28*60)/100

=2180/100

=21.8g/mol

6 0
1 year ago
How many grams of sulfur must be burned to give 100.0 g of So2
andriy [413]

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) →  SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

4 0
1 year ago
What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?
Ilya [14]

Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

4 0
1 year ago
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
1 year ago
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