How to balance this equation KClO4 → KCl + ?O2(g) and what type of reaction occurs?

Which of the following happens when an ionic bond is formed? (1 point)

bazaltina [42]

This problem is asking for an explanation of what happens when an ionic bond is formed. Although the choices are not given in the question, one can find them on the attached file and realize the answer is C "a less electronegative atom donates an electron to a more electronegative atom" according to:

<h3>Types of bonds:</h3><h3 />

In chemistry, the forces that hold atoms together are known as chemical bonds and act like connections for atoms to form compounds. There exist ionic and covalent bonds, so the formers occur when electrons are thoroughly donated from the least electronegative atom to the most electronegative one.

On the flip side, covalent bonds occur when the electrons are shared between the two or more of the atoms forming the compound. In such a way, one can discard choices A and B because they are more related to covalent bonds.

Therefore, one can select C "a less electronegative atom donates an electron to a more electronegative atom" as the correct answer, because not all the elements are able to donate more than one single electron, and the less its valency, the more ionic the compound turns out to be.

Learn more about types of bonds: brainly.com/question/792566

3 0

3 years ago

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

4 0

3 years ago

What is the trivial solution for a set of homogeneous equations? When is the solution not trivial. Give examples.

Monica [59]

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8 0

3 years ago

The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi

weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%$

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%$

So the answer to the second question is 26.90%.

3 0

3 years ago

As you go down group one of the periodic table, the reactions become and more

Lady bird [3.3K]

Answer:

vigorous

Explanation:

As you go down group one of the periodic table, the reactions become and more vigorous.

3 0

2 years ago