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sineoko [7]
2 years ago
7

Which of the following is the correct balanced equation for the reaction in which methane (CH4) burns in atmospheric oxygen (0₂)

to make carbon dioxide and water vapor?​
Chemistry
1 answer:
Talja [164]2 years ago
8 0

Answer: \text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}

Explanation:

The unbalanced equation is

\text{CH}_{4}+\text{O}_{2} \longrightarrow \text{CO}_{2}+\text{H}_{2}\text{O}

Balancing this equation, we get:

\boxed{\text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}}

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Which of these is an example of a wetland area? A. marsh B. stream C. city park D. farm land
Thepotemich [5.8K]
A.marsh it’s the most common one
5 0
4 years ago
Read 2 more answers
If you have 8.0 x 1025 molecules of water, how many moles is this?
Brilliant_brown [7]

Answer:

4.43

Explanation:

7 0
3 years ago
A gas mixture containing He, Ne, and Ar exerts a pressure of 3.00 atm. What is the partial pressure of each gas present in the m
matrenka [14]
<h3>Answer:</h3>

Partial pressure of He(P(He) = 1.5 atm.

Partial pressure of Ne(P(Ne) = 1 atm.

Partial pressure of Ar(P(Ar) = 0.5 atm.

<h3>Explanation:</h3>

According to Dalton law of partial pressure the sum of partial pressures of individual gases in a gaseous mixture is equivalent to the total pressure.

The partial pressure of a gas in a gaseous mixture is given by the product of the mole fraction and the total pressure.

Our gaseous mixture contains He, Ne, and Ar and the total pressure is 3 atm.

Since we are given the ratios of the gases in the mixture, we can calculate the partial pressure of each gas.

P(He) = 3/6 × 3 atm.

         = 1.5 atm.

P(Ne) = 2/6 × 3 atm.

        = 1 atm

P(Ar) = 1/6 × 3 atm.

        = 0.5 atm

Therefore, the partial pressures of gases He, Ne and Ar are 1.5 atm, 1 atm, and 0.5 atm respectively.

8 0
3 years ago
what is the indication that tells us what charge a transitional metal has since they dont have a set trend
Arlecino [84]

Answer:

Roman numbering in IUPAC naming system

Explanation:

This is quite an open question. Let's firstly separate the periodic table into two standard groups: group A elements and group B elements (transition metals).

The charge (or the oxidation state) of an element in group A can be identified by the group number. For example, group 1A elements would always have a charge of +1, as they have only one valence electron to lose.

Similar trend applies to group 2A: each element in that group would have a charge of +2, as each atom has 2 valence electrons to lose to become a cation.

You will notice that this is true fro group 3A and group 4A as well. Now, since an octet is the desired state for any species, starting with group 5A, it's easier to gain 3 electrons for species than lose 5 electrons to obtain an octer, meaning we'd expect -3 oxidation state for group 5A elements, -2 oxidation state for group 6A elements and -1 oxidation state for group 7A elements.

Notice that in the majority of cases, this is the standard trend and we'd generally only have one predominant oxidation state.

Considering group B, the transition metals, most of them have several oxidation states. That's why we usually memorize the ones which only have one oxidation state (such as zinc, silver) and in any other case when a transition metal has several oxidation states, they're identified in the name by using Roman numbering system.

Let's look at an example. Assume the problem states we have a salt which is iron chloride. This would be an improper name, as iron has two oxidation states: +2 and +3. That's why we have the rules of IUPAC naming to avoid ambiguity. If we had iron with an oxidation state of +2, we'd call the salt iron(II) chloride. An oxidation state of +3 would indicate iron(III) chloride.

To summarize, the main key of knowing the charge of a transition metal in a compound is to follow the IUPAC naming rules.

7 0
3 years ago
Is there a reaction between o2 and potassium iodide? If there is, please write it​
Soloha48 [4]

Answer:

Explanation:

Potassium iodide reacts with oxygen in presence of water to produce potassium hydroxide and potassium diiodoiodate(I) .

5KI + 2H₂O + O₂ =4 KOH + I₂ + K(II₂) .

7 0
3 years ago
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