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Ivan
3 years ago
6

A) The equilibrium shifts to the left, producing more H3O+.

Chemistry
1 answer:
Lunna [17]3 years ago
5 0

Answer: these are just options, what is the main question, without it, we cannot determine which option is correct, so please repost the question.

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Which choice lists the correct order of the coefficients of each substance in the following neutralization reaction when the equ
Gnoma [55]
The third one you have listed 


4 0
3 years ago
Given the following chemical reaction: 2 O2 + CH4 CO2 + 2 H2O What mass of oxygen is required to completely react with 20 grams
Jobisdone [24]
The chemical equation is:
CH₄ + 2O₂ → CO₂ + 2H₂O

First, we calculate the moles of methane present using:
Moles = mass / molecular mass
Moles = 20 / 16
Moles = 1.25

Next, we may observe from the chemical equation that the molar ratio between methane and oxygen is 1 : 2
So the moles of oxygen required are 2 x 1.25
2.5 moles of oxygen required

Mass = moles * molecular mass
Mass = 2.5 * 32
Moles = 80

C. 80 grams O₂
4 0
3 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
What are the 4 different types of chemical reactions? And how to remember them
Tju [1.3M]
<span>Synthesis, decomposition, single replacement and double replacement.</span>
4 0
3 years ago
Read 2 more answers
A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of
ICE Princess25 [194]

Answer:

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = V_1=10.0 L

Initial pressure of the gas = P_1=4.0 atm

Final volume of the gas = V_2=5.0 L

Final pressure of the gas = P_2=8.0 atm

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

PV=k

The equation given by this law is:

P_1V_1=P_2V_2

P_1\propto \frac{1}{V_1}

P_1V_1=k_i

k_i=4.0 atm\times 10.0 L = 40.0 atm L

P_2\propto \frac{1}{V_2}

P_V_2=k_f

k_f=8.0 atm\times 5.0 L = 40.0 atm L

k_i=k_f=40.0 atm L

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

6 0
3 years ago
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