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Ahat [919]
4 years ago
7

How could you verify that you produced carbon dioxide in your combustion reaction? 2. what indication did you have that nh3 was

produced in your decomposition reaction? 3. if you did not neutralize the saturated citric acid before you poured it down the drain describe the potential consequences to the plumbing and the environment?
Chemistry
1 answer:
DanielleElmas [232]4 years ago
5 0

1) Carbon dioxide is a gas, so when CO_{2} is evolved in the reaction, it appears as bubbles. The gas released extinguishes the fire and it can turn lime water milky.

Ca(OH)_{2} (aq) + CO_{2}(g) ----> CaCO_{3}(s) +H_{2}O(l)

2) When NH_{3} is released in a decomposition reaction we can identify by the strong pungent smell of the gas released.

3) Saturated citric acid can cause corrosion of the metal layers present in the pipes. So, before draining out any acid it is neutralized so that the pipes and other plumbing works do not get damaged leading to leaks in the drainage system.

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What combinations of proces can transform a metamorphic rock to sediments
grandymaker [24]
When metamorphic rocks are exposed to the earths surface they will be broken down into tiny fragments or pieces.Those are called sediments and after the tiny rocks are compacted against each other ,they will become sedimentary rocks .Hope this helped !
4 0
3 years ago
Which of these are true for the reaction above?
agasfer [191]

Answer:

a, d, f

Explanation:

ΔHrxn = ΔH(CCl4) -ΔH(CH4) = - 106.7 -(-74.8) = - 31.9 kJ/mol

6 0
4 years ago
What percent of an original sample remains after 3 half-lives? After 5 half-lives?
Salsk061 [2.6K]

Answer:

Three half lives corresponds to (12)3 . So a 18 quantity of the original isotope is retained. And the percentage of quantity of a radioactive material that remains after 5 half-lives will be . ∴NN0×100=10032=3.125.

7 0
4 years ago
When 16 g of methane (CH4) and 32 g of oxygen (O2) reacted to produce carbon dioxide and water, 11 g of carbon dioxide was produ
Aliun [14]

Answer:

Percent yield = 50%

Explanation:

Given data:

Mass of CH₄ = 16 g

Mass of O₂ = 32 g

Mass of CO₂ = 11 g

Percent yield of CO₂ = ?

Solution:

Chemical equation:

CH₄ + 2O₂    →  CO₂ + 2H₂O

Number of moles of CH₄:

Number of moles = mass/ molar mass

Number of moles = 16 g /16 g/mol

Number of moles = 1 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 32 g /32 g/mol

Number of moles = 1 mol

Now we will compare the moles of CO₂ with both reactant.

                             O₂             :            CO₂

                              2              :               1

                              1               :          1/2×1= 0.5 mol

                          CH₄              :            CO₂

                             1               :               1

Number of moles of CO₂ produced by oxygen are less so it will limiting reactant.

Theoretical yield:

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 0.5 mol × 44 g/mol

Mass = 22 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 11 g/ 22 g × 100

Percent yield = 50%

3 0
3 years ago
Explain the nucleus, electrons, protons, and neutrons of an atom.
kolezko [41]

Answer: I learned about this in school

Explanation:

Its very easy

8 0
3 years ago
Read 2 more answers
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