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andriy [413]
3 years ago
12

Can the process of rusting be called combustion? ​

Chemistry
2 answers:
gregori [183]3 years ago
8 0

Answer:

no the answer is oxidation

zavuch27 [327]3 years ago
4 0
The process of rusting cannot be called combustion due to the following reasons : ... No energy is obtained in a rusting process. Combustion is a chemical process in which a combustible substance reacts with oxygen to produce heat and light whereas in rusting, a metal reacts with oxygen to produce metallic oxide.
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Which statement is true for one molecule of sulfur trioxide?A.(There are three atoms of sulfur and one atom of oxygen.B.(There a
ki77a [65]
The answer would be D. This is because sulfur is on it's own, meaning one. while tri is a prefix for three so there are three oxygen atoms.
5 0
3 years ago
Read 2 more answers
How is coffee, Lemonade, Soda, Vinegar, Orange Juice an acid?
tatuchka [14]

Answer:

Curd, lemon juice, orange juice and vinegar taste sour. These substances taste sour because they contain acids. The chemical nature of such substances is acidic. The word acid comes from the Latin word acere which means sour.

8 0
3 years ago
Consider the following reaction.Cr2O3(s) + 3CCl4(l) 2CrCl3(s) + 3COCl2(g). When the green solid is mixed with the colorless liq
Elena L [17]

Answer:

The answer to your question is: letter B

Explanation:

Reaction

                 Cr2O3(s)   +   3CCl4(l)   ⇒  2CrCl3(s)  +   3COCl2(g)

From the information given and the reaction, we can conclude that:

Green solid = Cr2O3 (s)     "s" means solid

Colorless liquid = CCl4 (l)    "l" means liquid   and is the other reactant

Purple solid = CrCl3(s)        CrCl3 is purple and "s" solid

Then, as a green specks remains it means that the excess reactant is Cr2O3, so, CCl4 is the limiting reactant.

6 0
3 years ago
A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ
NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂   (1)        

b) The precipitate formed is Ni(OH)₂  

 

c) The limiting reactant is:

n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles

n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

n = \frac{2}{1}*0.030 moles = 0.060 moles                  

Hence, the limiting reactant is KOH.  

d) The mass of the precipitate formed is:

n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles

m = n*M = 0.010 moles*92.72 g/mol = 0.927 g  

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M  

C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M

C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M

I hope it helps you!                                                                        

5 0
3 years ago
An element that can have different physical properties or chemical arrangement is called a (n)
shusha [124]
<h3>Answer:</h3>

D. Allotrope

<h3>Explanation:</h3>

What is allotropy?

  • Allotropy refers to the existence of an element in more than one physical forms.
  • Allotropes are therefore different forms of an element with different physical properties or chemical arrangements.

What are some examples of allotropes?

  • Examples of elements that exhibit allotropy include, sulfur and carbon.
  • Allotropes of carbon are diamond and graphite.
  • Allotropes of sulfur are monoclinic sulfur and rhombic sulfur.
3 0
3 years ago
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