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3 years ago

How many single covalent bonds must a fluorine atom form to have a complete octet in its valence shell?

1 answer:

7 0

The answer to this question would be: 1

Fluorine is a nonmetal element that was located in the group 7 of the periodic table. Fluorine has 7 valence electron, only need 1 electron left to makes it fulfills the octet rule of 8 valence electron. One covalent bond should give fluorine the 1 electron it needs so the answer would be 1.

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ASAP MULTIPLE CHOICE WILL MARK BRAINLIEST

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the color of lines

Explanation:

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3 years ago

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The electron affinity of Bi is -94.6 kJ and the value for At is -280 kJ. What is the approximate electron affinity for Po?

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Answer:

48 KJ mol-1

Explanation:

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3 years ago

Olivia bought new gym shoes to play volleyball because she kept slipping when she ran in her old shoes. How will the new soles h

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Student Exploration: Limiting Reactants Vocabulary: chemical equation, chemical formula, chemical reaction, coefficient, limitin

SIZIF [17.4K]

Answer:

30 hot dogs

Explanation:

It is given that :

There are 4 packets of eight wieners, i.e. 4 x 8 = 32 wieners

There are 3 bags of ten buns, i.e. 3 x 10 = 30 buns

One hot dogs need 1 bun and 1 wiener to make a hot dog.

There are 30 buns, so 30 hot dogs can be made out by using all the 30 buns and the 30 wieners out of the 32 wieners.

Therefore, 30 hot dogs.

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3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago