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vova2212 [387]
3 years ago
8

. please help It's due today please

Chemistry
2 answers:
Anton [14]3 years ago
7 0

Answer:

??????????

Explanation:

gtnhenbr [62]3 years ago
3 0

Answer:

i have no clue

Explanation:

You might be interested in
Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g CO2 and 0.6551 g
Ede4ka [16]

Answer:

C2H4O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.6004/44 = 0.0364

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0364= 0.4368g

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.6551/18 = 0.0364 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0364= 0.0728

The mass of hydrogen is thus 0.0728* 1 = 0.0728g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.8009 - 0.0728 - 0.4368 = 0.2913 mole

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.2913/16 = 0.0182 mole

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon and oxygen 0.0182

H = 0.0728/0.0182 = 4

C= 0.0364/0.0182 = 2

O= 0.0182/0.0182= 1

The empirical formula is thus C2H4O

3 0
3 years ago
Consider the reaction: 2A(g)+B(g)→3C(g). When A is changing at a rate of -0.110M⋅s−1, How fast is C increasing?
mr_godi [17]
From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
hence dB / dt = -0.110 / 2 = -0.055 M/s
dC / dt = (-3/2) (-0.110) = 0.165 M/s 
5 0
3 years ago
Octasulfur reacts with pure oxygen to produce sulfur dioxide. Determine the percent
Margaret [11]

Answer:

Octasulfur is just S8. Eight S atoms in

a sort of crown shape. Sulfur Dioxide is a gas, SO2. Does that help?

Explanation:

5 0
3 years ago
Be sure to answer all parts.
Fudgin [204]

Answer:

2H₂ + O₂       →     2H₂O

Explanation:

Chemical equation:

H₂ + O₂       →     H₂O

Balance chemical equation:

2H₂ + O₂       →     2H₂O

Step 1:

H₂ + O₂       →     H₂O

Left hand side                     Right hand side

H = 2                                    H = 2

O = 2                                    O = 1

Step 2:

H₂ + O₂       →     2H₂O

Left hand side                     Right hand side

H = 2                                    H = 4

O = 2                                    O = 2

Step 3:

2H₂ + O₂       →     2H₂O

Left hand side                     Right hand side

H = 4                                    H = 4

O = 2                                    O = 2

7 0
3 years ago
A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent compos
Rama09 [41]

Answer:

53.85%

Explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%

3 0
3 years ago
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