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Digiron [165]
3 years ago
12

Identify the combustion reaction. identify the combustion reaction. c(s) + o2(g) → co2(g) 2 h2(g) + o2(g) → 2 h2o(g) c3h8(g) + 5

o2(g) → 3 co2(g) + 4 h2o(g) 2 c3h7oh(l) + 9 o2(g) → 6 co2(g) + 8 h2o(g) all of the reactions are examples of combustion reactions.
Chemistry
1 answer:
tester [92]3 years ago
7 0
Combustion reaction occurs when organic compound reacts with oxygen to form CO₂, H₂O and energy
C(s) + O₂(g) → CO₂(g) is Synthesis reaction not combustion
2 H₂(g) + O₂(g) → 2 H₂O(g) also Synthesis reaction and not combustion
C₃H₈(g) + 5 O₂ → 3 CO₂(g) + 4 H₂O(g) is considered as Combustion reaction
2 C₃H₇OH(l) + 9 O₂(g) → 6 CO₂(g) + 8 H₂O(g) Combustion reaction
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The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.
exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

8 0
3 years ago
Help please
Lady_Fox [76]

Answer:

It is this below

Explanation:

P = m/v

8 0
2 years ago
Given that the ka for hocl is 3.5 × 10–8, calculate the k value for the reaction of hocl with oh–.
Harlamova29_29 [7]
Following reaction is involved in above system
HOCl(aq)  ↔  H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l)  ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K  = [OCl-][H+<span>]/[HOCl]   ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>)   ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K  = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14

</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these
6 0
3 years ago
Read 2 more answers
water is 11.19% hydrogen and 88.1% oxygen by mass. Lake superior contains about 1.2x10 kg of water. what mass of hydrogen is con
KonstantinChe [14]

Answer:

mass of hydrogen in lake = 11.19% of 12

                                          = 1.3428

check if question is correct or not

Explanation:

3 0
2 years ago
Read 2 more answers
What is the limiting reactant when 4 mol P4 and 4 mol S8 react.
In-s [12.5K]

Taking into account the reaction stoichiometry, P₄ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

8 P₄ + 3 S₈ → 8 P₄S₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • P₄: 8 moles
  • S₈: 3 moles
  • P₄S₃: 8 moles

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 3 moles of S₈ reacts with 8 moles of P₄, 4 moles of S₈ reacts with how many moles of P₄?

moles of P_{4} =\frac{4 moles of S_{8} x8 moles of P_{4} }{3 moles of S_{8}}

<u><em>moles of P₄= 10.667 moles</em></u>

But 10.667 moles of P₄ are not available, 4 moles are available. Since you have less amount of moles than you need to react with 4 moles of S₈, P₄ will be the limiting reagent.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
1 year ago
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