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vitfil [10]
3 years ago
15

MATCH THESE ITEMS

Chemistry
2 answers:
butalik [34]3 years ago
6 0
In the photo……………………….

agasfer [191]3 years ago
4 0

Answer:

Picture

Explanation:

Chemistry-

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How many oxygen atoms are in the reactants?
Pani-rosa [81]
The answer would be 2
7 0
3 years ago
A pure element unbound or in a diatomic state, such as cl2, always has an oxidation number of ________. a. 1 b. 2 c. magnitude e
nignag [31]

A pure element unbound or in a diatomic state, such as cl2, always has an oxidation number of 0 (zero).

<h3>Why does pure element or a diatomic molecule has zero oxidation state?</h3>

In a neutral substance with atoms of only one element, the oxidation number of an atom is zero. As a result, the oxidation number of the atoms in O2, O3, P4, S8, and aluminum metal is 0. The oxidation numbers for an element in its normal state will be zero. O2 and Cl2 are diatomic gas molecules that occur naturally, thus when they are in that state, they have an oxidation state of zero. Metals like zinc will also have an oxidation number of zero if they are in their natural solid state.

O2 and Cl2 are neutral diatomic, hence they will always have a zero oxidation state. It is impossible for one oxygen atom to have a negative 2 charge while the other has a positive 2. The oxidation states should be 0 if the elements are solids, liquids, or any type of diatomic molecule.

Learn more about oxidation state here:

brainly.com/question/6707068

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8 0
2 years ago
Read 2 more answers
State two properties of sulphur that makes it possible to extract using the frasch process​
Wewaii [24]

Explanation:

sulphur has a low melting point.

sulphur has a low boiling point thus it can be extracted by pumping hot water which turns it ito solution form

8 0
2 years ago
What is the gram of hydrogen in periodic table?<br>​
earnstyle [38]

Answer:

This is your answer

Hope it helps!!!

5 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
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