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3 years ago

A bond formed when atoms transfer elentron is a bond

1 answer:

5 0

Ionic bonds involve a cation and an anion. The bond is formed when an atom, typically a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to acquire the electron(s) to become a negative ion, or anion.

One example of an ionic bond is the formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom. In this reaction, the sodium atom loses its single valence electron to the fluorine atom, which has just enough space to accept it. The ions produced are oppositely charged and are attracted to one another due to electrostatic forces.

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g This plot shows the rate of the decomposition of SO2Cl2 into SO2 and Cl2 as a function of the concentration of SO2Cl2. What is

scoundrel [369]

Answer:

When n = 1, the reaction is of the First Order

Explanation:

Find attach the solution

5 0

3 years ago

2. Consider the reaction 2 Cg H18 (4) +250â (9) âº 16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1

Sladkaya [172]

Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of $C_8H_{18}$ = 16.15 moles

First we have to calculate the moles of $H_2O$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react to give 18 moles of $H_2O$

So, 16.15 moles of $C_8H_{18}$ react to give $\frac{16.15}{2}\times 18=145.35$ moles of $H_2O$

The moles of water produced are 145.35 moles.

<u>Solution for part (b) : Given,</u>

Mass of $C_8H_{18}$ = 878 g

Molar mass of $C_8H_{18}$ = 114 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_8H_{18}$ .

$\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react with 25 moles of $O_2$

So, 7.702 moles of $C_8H_{18}$ react with $\frac{7.702}{2}\times 25=96.275$ moles of $O_2$

Now we have to calculate the mass of $O_2$ .

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g$

The mass of oxygen needed are 3080.8 grams.

6 0

2 years ago

In a reaction of a potential new fuel, it is found that when 2.81 moles of the fuel combusts, 1,612 kJ of energy is released. Wh

Tema [17]

Answer:

-573.67

Explanation:

whenever energy is released in a chemical reaction, we would then expect the delta H of the reaction to be negative because the reaction is an exothermic reaction.

now we have that 2.81 moles of fuel when it combusts would releases 1612kJ of energy

thus, 1 mole will release 1612/2.81 = -573.67kJ of heat

Therefore the delta H of the reaction = -573.67 kJ/mol

3 0

2 years ago

The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product

Darya [45]

Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

k = rate constant = $0.610M^{-1}s^{-1}$

$a_0$ = initial concentration = 0.860 M

a= concentration left after time t = 0.260 M

$\frac{1}{0.260}=0.860\times t+\frac{1}{0.860}$

$t=3.120s$

Thus it takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

8 0

3 years ago

1. Express in conventional notation (no exponents) in the space provided within the

mina [271]

Answer:

a) 320: two significant figures.

b) 2,366: four significant figures.

c) 73.0: three significant figures.

d. 532.5: four significant figures.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to write each number by knowing we move the decimal places to the right as much as the exponent is, and also, we count every figure, even zeros, because they are to the right of the first nonzero digit:

a) 320: two significant figures because the rightmost zero is not preceded o followed by a decimal place.

b) 2,366: four significant figures.

c) 73.0: three significant figures, because the zero is followed by the decimal place.

d. 532.5: four significant figures.

Regards!

7 0

3 years ago