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Ionic bonds are formed between metal and nonmetal elements. Potassium has only one valence electron and it is stripped away in t

lbvjy [14]

Refer to the table below. Credits to https://terpconnect.umd.edu/~wbreslyn/chemistry/naming/IonicCharge2.jpg

Cations with (+ ) charges lose electrons in order to obtain an octet (8 valence electrons) when they ionically bond with another ion. We're looking for the ions that loses electrons here. So, from the table:

Al 3+ , S 2- , O 2-, Ag + , Ne ( noble gas, no charge)

Since Al and Ag has (+) charges, they are going to lose electrons to form ionic bonds with other atoms.

5 0

3 years ago

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How many molecules of water are in 3.8 moles of water?

Salsk061 [2.6K]

Answer:

2.29*10^24 molecules of water

Explanation:

1 molof water =6.022*10^23 molecules of water

therefore

molecules of water = 3.8*6.022*10^23

5 0

2 years ago

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For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when

KATRIN_1 [288]

Answer: Concentration of N₂ is 4.8. $10^{9}$ M.

Explanation: $K_{c}$ is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For

N2(g) + 2 O2(g) ⇄ 2 NO2(g)

$K_{c}$ = $\frac{[NO2]^{2} }{[N2][O2]^{2} }$

From the question concentration of NO2 is twice of O2:

[NO2] = 2[O2]

Substituting this into $K_{c}$ :

$K_{c}$ = $\frac{[2O2]^{2} }{[N2][O2]^{2} }$

8.3. $10^{-10}$ = $\frac{4O2^{2} }{[N2].O2^{2} }$

[N2] = $\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }$

[N2] = $\frac{4}{8.3.10^{-10} }$

[N2] = 4.8. $10^{9}$

The concentration of N2 in the equilibrium is [N2] = 4.8. $10^{9}$ M.

6 0

3 years ago

Which element has chemical properties that are most similar to the chemical properties of flourine

aleksandrvk [35]

The elements in group 7, which fluorine is in also, would have similar properties due to the valence electrons and the configuration.

3 0

2 years ago

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Need help with 22 and 24

choli [55]

Answer:

22:

Formular:

$atomic \: mass = \frac{ \sum(isotopic \: mass \times \%abundance)}{100} \\$

substitute:

$atomic \: mass = \frac{(23.985 \times 78.70) + (24.986 \times 10.13) + (25.983 \times 11.17)}{100} \\ \\ = \frac{(1887.620) + (253.108) + (290.230)}{100} \\ \\ = \frac{2430.958}{100} \\ \\ { \boxed{ \boxed{average \: atomic \: mass = 24.3 \: amu}}}$

23:

Same element is represented by same number of protons. 

Answer:

6 protons. 6 protons

7 neutrons. 8 neutrons

6 electrons. 6 electrons

Note: Atoms with same proton number but different mass number are called isotopes

5 0

2 years ago

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