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GaryK [48]
3 years ago
5

Use dhe References to access inportaat valbes ifneeded for ais A sample of 2.601 grams of a compound containing carbon and hydro

gen reacts with oxygen at elevated temperatures to yield 8.793 grams of CO2 and 1.800 grams of H2O. (a) Calculate the masses of C and H in the sample Grams: C- H- (b) Does the compound contain any other elements? (e) What are the mass percentages of C and H in the compound? Mass percentages: C % ; H-| (d) What is the empirical formula of the compound? Esler the elements in the order presested in the question empirical formula-
Chemistry
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

jhouhglhi

Explanation:

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In the formula for carbon tetrahydride, CH4, the percent composition of all elements is:
creativ13 [48]
C%=12/16*100=3/4*100=75%
H%=25%
8 0
3 years ago
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What are twin primes between 1 to 100<br>​
Furkat [3]
The twin primes between 1 and 100 are; (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73). Hope this helped<3
6 0
3 years ago
the decomposition of calcium carbonate to calcium oxide and carbon dioxide only takes place at very high temperatures, making th
USPshnik [31]
So calculate the H for the other two reactions a room temperature and combine the reactions to calculate the H of the decomposition of calcium carbonate using the Hess's Law
4 0
3 years ago
How many grams of magnesium acetate are in 8.95x10^23 formula units?
Natasha_Volkova [10]

Answer:

211.63 g.

Explanation:

  • Particles could refer to atoms, molecules, formula units.
  • <em>Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).</em>

<em><u>Using cross multiplication:  </u></em>

1.0 mole → 6.022 x 10²³ molecules.

??? mole → 8.95 x 10²³ molecules.

  • The no. of moles of magnesium acetate = (8.95 x 10²³ molecules) (1.0 mole) / (6.022 x 10²³ molecules) = 1.486 mol.

∴ The grams of magnesium acetate are in 8.95 x 10²³ formula units = n x molar mass = (1.486 mol)(142.394 g/mol) = 211.63 g.

5 0
3 years ago
How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut
san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

<em>Where X is the amount of material in grams that comes to the organic phase.</em>

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

<em>In the second extraction, you will extract 0.50g of material</em>

Thus, after the two extraction you will lost:

4.44g + 0.50g = <em>4.94g of material</em>

<em></em>

6 0
3 years ago
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