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andreyandreev [35.5K]

3 years ago

9

Which atom or ion is the largest? A. K B. K+ C. Ca D. Ca2+ E. Li

1 answer:

pishuonlain [190]3 years ago

5 0

Answer:

A. K

Step-by-step explanation:

Remember the trends in the Periodic Table:

Atomic radii <em>decrease</em> from left to right across a Period.
Atomic radii <em>increase</em> from top to bottom in a Group.
Ionic radii of metal cations are <em>smaller</em> than those of their atoms.

Thus, the largest atoms are in the lower left corner of the Periodic Table.

The diagram below shows that K is closest to the lower left, so it is the largest atom. It is also larger than any of the cations.

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Give the spectator ions for the reaction that occurs when aqueous solutions of H 2SO 4 and KOH are mixed.

Sever21 [200]

Answer:

The spectator ions is: $K^+$ and $SO^{2-}_4$

Explanation:

The equation of reaction between H₂ SO₄ and KOH is:

$H_2SO_{4(aq)} + 2KOH _{aq} \to K_2SO_{4(aq)} +2H_2O _{(l)}$

Rewriting this equation as ionic;

$[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+ SO_4^{2-} + 2H_2O ]$

Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.

5 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

How is n1 in the Rydberg equation related to the quantum number n in the Bohr model?How is n1 in the Rydberg equation related to

andre [41]

Ansddwdccwer:

ExplanationBecause:

7 0

4 years ago

True or False ... In a chemical change, a new<br> substance is formed.

nikitadnepr [17]

The answer is true the answer has to be 20 words long so true true true true true true

7 0

3 years ago

A solution of sulfuric acid has a concentration of 0.0980 g/L. If the density of the acid is 1.84 g/mL, what is the concentratio

mariarad [96]

The answer is 98ppm.

The ppm (Parts per million) is also a concentration unit. 1 ppm is equivalent to 1mg/L
Now, concentration of the solution of sulfuric acid is 0.0980 g/L.
We rewrite, 0.0980 g = 0.0980*1000 mg = 98mg
Therefore, the concentration of the sulfuric acid solution is 98 mg/L = 98 ppm.

6 0

4 years ago

Read 2 more answers