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schepotkina [342]
3 years ago
10

Which is an example of a chemical reaction?

Chemistry
1 answer:
allochka39001 [22]3 years ago
5 0
No it's B because of a chemical reaction between the air and iron
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Why do we have 2 high tides and 2 low tides a day?
Sunny_sXe [5.5K]
Whereas semidiurnal tides are observed at the equator at all times, most locations north or south of the equator experiencetwo unequal high tides and twounequal low tides per tidal day; this is called a mixed tide and the difference in height between successive high (or low) tides iscalled the diurnal inequality.
5 0
3 years ago
How many molecules are there in 6.73 moles of phosphorus trichloride? (write your answer in scientific notation).
anyanavicka [17]

Answer: There are 4.05\times 10^{24}  molecules in 6.73 moles of phosphorus trichloride

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of PCl_3 contains = 6.023\times 10^{23}  molecules

Thus 6.73 moles of PCl_3 contains = \frac{6.023\times 10^{23}}{1}\times 6.73=4.05\times 10^{24}  molecules

Thus there are 4.05\times 10^{24}  molecules in 6.73 moles of phosphorus trichloride

3 0
3 years ago
In Europe, nutritional information is given in kilojoules (kJ) instead of nutritional calories (1 nutritional calorie = 1 kcal).
Alinara [238K]
There would be 68 calories per cup
8 0
2 years ago
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Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of h2c2o4 and h2o in th
shusha [124]

Answer:

See explanation below

Explanation:

The overall reaction, is:

MnO4⁻(aq) + H₂C₂O₄(aq) ---------> Mn²⁺(aq) + CO₂(g)

Balancing this redox reaction means that one compound is reducting while the other is oxidizing. So, we need to separate both compounds into 2 semi equations and balance both of them, per separate and then, we can join them.

As we want to balance in acid medium, means that we need to add water and H⁺ in both reactions. Doing that we have the following:

MnO₄⁻   ---------------> Mn²⁺

In this reaction, we can clearly see that it's not balanced. To balance this semi equation, let's see the elements. In the reactans we have Mn and O, but in the products we only have Mn, the atom of oxygen where could it be? As we are doing acid medium, if in the reactants we have oxygen, this oxygen can be as products in the form of water, so we add water there.

MnO₄⁻   ---------------> Mn²⁺ + H₂O

Now, the water has hydrogen atoms, and if we are in acid medium, the hydrogen can only come from the acid medium, and in this case H⁺ so:

H⁺ + MnO₄⁻ ----------> Mn²⁺ + H₂O

Now, it's time to balance the charges. First Mn²⁺ is the lowest oxidation state of the manganese, this means that in the reactants Mn is passing from a higher state to a lower state, therefore, this compouns is reducting. How many electrons? well, in this case, we know that oxygen usually have the oxidation state -2, so the manganese would be:

(-2 * 4) + x = -1

-8 + x = -1 -------> x = +7

Therefore, manganese passes from 7+ to 2+, it's gaining 5 electrons so:

H⁺ + MnO₄⁻ + 5e⁻ ----------> Mn²⁺ + H₂O

Finally, we just balance the masses and charges:

8H⁺ + MnO₄⁻ + 5e⁻----------> Mn²⁺ + 4H₂O

Now, we just do the same thing with the other semi equation which is oxydizing. The explanation of that, is similar to this, so I'm gonna do it directly:

C₂O₄²⁻ -----------> CO₂

In this case, we can easily see that carbon is losing 2 electrons, so, let's put the 2 electrons on the product to balance the charges, and then, the masses:

C₂O₄²⁻ -----------> CO₂ + 2e⁻

C₂O₄²⁻ -----------> 2CO₂ + 2e⁻

Let's join both equations and do the sum of them:

8H⁺ + MnO₄⁻ + 5e⁻----------> Mn²⁺ + 4H₂O

C₂O₄²⁻ -----------> 2CO₂ + 2e⁻

As we can see, we do not have the same electrons on both equations, we need to equal those values so:

2 * (8H⁺ + MnO₄⁻ + 5e⁻----------> Mn²⁺ + 4H₂O)

5 * (C₂O₄²⁻ -----------> 2CO₂ + 2e⁻)

16H⁺ + 2MnO₄⁻ + 10e⁻----------> 2Mn²⁺ + 8H₂O

5C₂O₄²⁻ -----------> 10CO₂ + 10e⁻

Now, let's sum both equations:

16H⁺ + 2MnO₄⁻ + 10e⁻----------> 2Mn²⁺ + 8H₂O

5C₂O₄²⁻ -----------> 10CO₂ + 10e⁻

___________________________________

16H⁺ + 2MnO₄⁻ + 5C₂O₄²⁻ -----------> 2Mn²⁺ + 10CO₂ + 8H₂O

This would be the balanced reaction, however, let's put it as it was originally with the H2 in the C2O4 and balance it:

<h2>2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ -----------> 2Mn²⁺ + 10CO₂ + 8H₂O</h2>
7 0
3 years ago
Use the periodic table to classify each of the elements below.
Alika [10]

Answer:

Explanation:

Cadmium:(Cd)

Cadmium is transition metal present in group twelve. It is soft metal and properties are similar to the other group members like zinc and mercury. Its atomic number is forty eight and have two valance electrons.

Electronic configuration:

Cd₄₈ = [Kr] 4d¹⁰ 5s²

Vanadium: (V)

It is present in group five. It is malleable and ductile transition metal. Its atomic number is twenty three. Vanadium have five valance electrons.

Electronic configuration:

V₂₃ =[Ar] 3d³ 4s²

Xenon: Xe

Xenon is present in group eighteen. It is noble gas. Its outer most valance shell is complete that's why it is inert. its atomic number is fifty four. Xenon have eight valance electrons.

Electronic configuration:

Xe₅₄ = [Kr] 4d¹⁰ 5s² 5p⁶

Iodine: (I)

Iodine is present in group seventeen. Its outer most valance shell have seven electrons. Iodine is the member of halogen family. It gain one electron to complete the octet. its atomic number is fifty three.

Electronic configuration:

I₅₃ = [Kr] 4d¹⁰ 5s² 5p⁵

Potassium: (K)

Potassium is present in group one. it is alkali metal. Its atomic number is nineteen. Its valance shell has one electron. Potassium loses its one valance electron and gets stable electronic configuration.

Electronic configuration:

K₁₉ = [Ar] 4s¹

Strontium: Sr

Strontium is present in group two. it is alkaline earth metal. its atomic number is thirty eight and have two valance electrons.

Electronic configuration:

Sr₃₈ = [Kr] 5s²

5 0
3 years ago
Read 2 more answers
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