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Gwar [14]
3 years ago
5

The unbalanced equation below shows the combustion of methane.

Chemistry
1 answer:
Ratling [72]3 years ago
3 0
B. 1, 2, 1, 2. You balance the equation by making sure there are equal elements on both sides.
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Which isomer of 1-bromo-2-tert-butylcyclohexane reacts faster when refluxed with potassium tert-butoxide?Draw the structure of t
ivolga24 [154]

Answer:

See explanation and image attached

Explanation:

The reaction of 1-bromo-2-tert-butylcyclohexane with  potassium tert-butoxide is an elimination reaction that occurs by E2 mechanism.

The E2 reaction proceeds faster when the hydrogens are in an antiperiplanar position at an angle of 180 degrees.

This is only attainable in the trans isomer of 1-bromo-2-tert-butylcyclohexane. Hence trans 1-bromo-2-tert-butylcyclohexane reacts faster with potassium tert-butoxide

8 0
3 years ago
There are 100.0 grams of each reactant available determine the limiting reactant in this equation
Romashka [77]
Since you have not included the chemical reaction I will explain you in detail.

1) To determine the limiting agent you need two things:

- the balanced chemical equation

- the amount of every reactant involved as per the chemical equation

2) The work is:

- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.


- determine the number of moles of each reactant with this formula:

number of moles = (mass in grams) / (molar mass)

- set the proportion with the two ratios (theoretical moles and actual moles)


- compare which reactant is below than the stated by the theoretical ratio.

3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:

i) Chemical equation: H₂ + O₂ → H₂O

ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O

iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂

iv) Covert 100 g of H₂ into number of moles

n = 100g / 2g/mol = 50 mol of H₂

v) Convert 100 g of O₂ to moles: 

n = 100 g / 32 g/mol = 3.125 mol

vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂

vii) Compare the two ratios:

2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂

Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.

Therefore, the O₂ is the limiting reactant in this example.

7 0
3 years ago
Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu
ad-work [718]

Initial volume of the balloon = V_{1}= 348 mL

Initial temperature of the balloon T_{1} = 25.0^{0}C + 273 = 298 K

Final volume of the balloon V_{2} = 322 mL

Final temperature of the balloon = T_{2} = ?

According to Charles law, volume of an ideal gas is directly proportional to the temperature at constant pressure.

\frac{V_{1} }{T_{1} } =\frac{V_{2} }{T_{2} }

On plugging in the values,

\frac{348mL}{298 K} =\frac{322 mL}{T_{2} }

T_{2} =276 K

Therefore, the temperature of the freezer is 276 K

5 0
3 years ago
What is the pH of a solution with an [H+] of (a) 5.4 x 10-10, (b) 4.3 x 10-5, (c) 5.4 x 10-7?
IgorLugansk [536]

Answer:

a. 9.2

b. 4.4

c. 6.3

Explanation:

In order to calculate the pH of each solution, we will use the definition of pH.

pH = -log [H⁺]

(a) [H⁺] = 5.4 × 10⁻¹⁰ M

pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2

Since pH > 7, the solution is basic.

(b) [H⁺] = 4.3 × 10⁻⁵ M

pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4

Since pH < 7, the solution is acid.

(c) [H⁺] = 5.4 × 10⁻⁷ M

pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3

Since pH < 7, the solution is acid.

3 0
3 years ago
We can consider a chemical to be safe if:
slamgirl [31]
We can consider a chemical to be safe if it does not contain any harmful substances such as nitrogen, harmful acids, or even excessive heat. Other chemicals such as the chemicals used in food products like, for example, citric acid, is not so harmful.
Hope this helps! :D
Any questions? Just let me know! I'd be happy to help any way possible.
8 0
3 years ago
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