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choli [55]
3 years ago
12

What type of chemical bond would form between an atom of lithium (Li) and an atom of chlorine (Cl). Explain specifically why thi

s type of bond would form.
Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
7 0

Explanation:

  • When a bond is formed by transfer of electrons from one atom to another then it results in the formation of an ionic bond.

An ionic bond is generally formed by a metal and a non-metal.

For example, lithium is an alkali metal with atomic number 3 and its electronic distribution is 2, 1.

And, chlorine is a non-metal with atomic number 17 and its electronic distribution is 2, 8, 7.

So, in order to complete their octet lithium needs to lose an electron and chlorine needs to gain an electron.

Hence, both of then on chemically combining together results in the formation of an ionic compound that is, lithium chloride (LiCl).

An ionic compound is formed by LiCl because lithium has donated its valence electron to the chlorine atom.  

  • On the other hand, if a bond is formed by sharing of electrons between the two chemically combining atoms then it is known as a covalent bond.

For example, O_{2} is a covalent compound as electrons are being shared by each oxygen atom.

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Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17
Hitman42 [59]

<span>Answer is: mass of burned butane is 11.6 g.</span>

Chemical reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

m(butane) = 50,0 g.

<span> V(CO</span>₂) = 17,9 L.<span>
n(CO</span>₂) = V(CO₂) ÷ Vm.<span>
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n(CO</span>₂) = 0,8 mol.<span>
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5 0
3 years ago
Initially, a particular sample has a total mass of 200 grams and contains 128 x 1010 radioactive nuclei. These radioactive nucle
lubasha [3.4K]

Explanation:

Formula to calculate how many particles are left is as follows.

              N = N_{0} (\frac{1}{2})^{l}

where,     N_{0} = number of initial particles

                              l = number of half lives

As it is given that number of initial particles is 128 \times 10^{10} and number of half-lives is 3.

Hence, putting the given values into the above equation as follows.

               N = N_{0} (\frac{1}{2})^{l}

                    = 128 \times 10^{10}(\frac{1}{2})^{3}

                    = 16 \times 10^{10}

or,                    1.6 \times 10^{11}

Thus, we can conclude that 1.6 \times 10^{11} particles of radioactive nuclei remain in the given sample.

In five hours we've gone through 5 half lives so the answer is:

particles

8 0
3 years ago
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