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ANEK [815]
3 years ago
9

The following question was posed on an exam:

Chemistry
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

The answer given by the student is not totally correct.

Explanation:

Law of multiple proportions states that an element Q will react with different volume of Fluorine to produce two non-similar compounds. Hence, the ratio of the masses(Fluorine only) needs to be an absolute reduced value.

Given that:

3.2 g of a sample of Q reacts with Fluorine to form 10.8 g of the unknown fluoride A.

This means the mass of Fluorine present in the compound = 10.8 g - 3.2 g = 7.6 g

Thus, 3.2 g of a sample of Q reacts with 7.6 g of Fluorine.

Also, 6.4 g sample of Q reacts with Fluorine to form 29.2 g of unknown fluoride B.

If we divide the samples by (2), we have 3.2 g sample of Q reacting with Fluorine to form 14.6 g of unknown fluoride B.

This means the mass of Fluorine present in the compound = 14.6 g - 3.2 g = 11.4 g

Thus, 3.2 g Q reacted here with 11.4 g fluorine.

Hence, 11.4/7.6 = 1.5 = 3/2

This above therefore satisfies the law of multiple proportions.

This is not aligned with what the student did, what the student did was to relate the amount of Q used to make A and B. Suppose, we start with twice amount of Q, the ratio would have been smaller(i.e. the ratio of Q).

So, this doesn't relate to the law of multiple proportions.

The law of multiple proportions is specifically concerned with the mass of the element rather than Q that can react with Q. Therefore, there is more reason to relate the two samples of equal masses of Q that react with different masses of Fluorine. The ratios of Fluorine will then be small whole numbers.

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How many moles of oxygen are required to react completely with 5 mol C8H18?
Nata [24]

Answer:

62.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.

Thus, 62.5 moles of O₂ is needed for the reaction.

6 0
3 years ago
HELP HURRY‼️‼️‼️<br> FAST
lesya692 [45]

I believe the answer is D

8 0
2 years ago
Read 2 more answers
2c+02=2CO2. The moles of co2 produced when 0.25 moles of O2 react is?​
Sophie [7]
<h3>Answer:</h3>

\displaystyle 0.5 \ mol \ CO_2

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Moles
  • Compounds

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C + O₂ → 2CO₂

[Given] 0.25 moles O₂

[Solve] moles CO₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol O₂ → 2 mol CO₂

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.25 \ moles \ O_2(\frac{2 \ mol \ CO_2}{1 \ mol \ O_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                        \displaystyle 0.5 \ mol \ CO_2
7 0
2 years ago
LOTS OF POINTS PLZ HELP!!!!!<br><br><br> ...I got the first two for ya...
vaieri [72.5K]
I believe this is it

7 0
3 years ago
Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH
uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

7 0
3 years ago
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