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nexus9112 [7]
3 years ago
12

in terms of erosion, which letter displays the word meander? ( A.) W (B.) F ( C.) A ( D.) S

Chemistry
1 answer:
nikklg [1K]3 years ago
8 0
I believe it is answer D
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Why are the alkali metals and the halogens very reactive
jeyben [28]

Answer:

Answer in explanation

Explanation:

The reactivity or passiveness of an element depends solely on how close it is to attain a noble gas configuration. This means the closer an element is to attain a noble gas configuration, the greater its reactivity in both direction, positively or negatively.

Alkali metals belong to group 1 of the periodic table while halogens belong to group 17 of the periodic table. This means they are just one electron away from achieving the stability of a noble gas configuration. While alkali metals need to lose one electron to form a univalent positive ion, halogens news to gain one electron to form a univalent negative ion.

They tend to go about this vigorously and as such undergo several chemical reactions because of that single electron they neeed.

4 0
2 years ago
copper hydroxide and potassium sulfate are produced when potassium hydroxide reacts with copper sulfate balanced equation
a_sh-v [17]

This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.

<h3>Balancing chemical equations:</h3>

In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.

For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:

KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2

As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:

2KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2

Learn more about balancing chemical equations: brainly.com/question/8062886

8 0
2 years ago
Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su
Ratling [72]

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

<em>1</em> C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • C₅H₁₂ is the only <em>reactant</em> that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
  • CO₂ is the only <em>product</em> that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

<em>1 </em>C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + ? H₂O

  • Similarly, C₅H₁₂ is the only <em>reactant</em> that contains H atoms. H₂O is the only <em>product</em> that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

<em>1</em> C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

  • Both CO₂ and H₂O are <em>products</em> that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the <em>product</em> side corresponds to 8 O₂ molecules on the reactant side.

<em>1</em> C₅H₁₂ + <em>8</em> O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

  • All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", <em>GCSE Chemistry (Single Science)</em>, BBC Bitesize.

5 0
3 years ago
23. The cell has a
andrew11 [14]
Then that would be 100% all together I believe
5 0
1 year ago
While a certain isotope decays, it emits photons. What kind of decay is happening?
gizmo_the_mogwai [7]
The correct answer is Gamma decay. It happens after beta and alpha decay because what's left after those two can enter a new process of gamma decay. This releases gamma rays which is a more complex term for the photons that you mentioned before. These rays can be dangerous for humans so care not to get caught in them.
6 0
3 years ago
Read 2 more answers
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