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Westkost [7]
3 years ago
13

What are the trends for atomic size and ionization? Explain why for each trend. Tell me the direction- up or down a group and ac

ross a period for each trend. Use the lecture, the video at the end of the lecture, or the guide for periodic trends to respond to this prompt.
Chemistry
1 answer:
Alinara [238K]3 years ago
5 0

Answer:

See explanation

Explanation:

Atomic size increases down the group due to the addition of more shells.

As more shells are added and repulsion of inner electrons become more significant, atomic size increases down the group. However, across the period, atomic size decreases due to increase in effective nuclear charge without any increase in the number of shells. This causes increased attraction between the nucleus and the outermost shell thereby decreasing the size of the atom.

Ionization energy decreases down the group because the outermost electron is more shielded by inner electrons making it easier for this outermost electron to be lost. Across the period, ionization energy increases due to increase in effective nuclear charge which makes it more difficult to remove the outermost electron due to increased nuclear attraction.

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A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and
Dmitry_Shevchenko [17]

Answer:

1. Rate =k [NO]^{2}[Cl_{2}]

2. k= 0.42 \frac{L^{2}}{mol^{2}*s}

Explanation:

Rate =k [NO]^{m}[Cl_{2}]^{n}

Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2

Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1

Rate =k [NO]^{2}[Cl_{2}]^{1}

Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12}  = 0.42 \frac{L^{2}}{mol^{2}*s}

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Compared to oil and gas, solar energy:
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Answer:

a. 1.12 L

Explanation:

Step 1: Write the balanced equation for the photosynthesis

6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)

Step 2: Calculate the moles corresponding to 2.20 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

2.20 g × 1 mol/44.01 g = 0.0500 mol

Step 3: Calculate the moles of O₂ produced

The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol

Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP

At STP, 1 mole of O₂ occupies 22.4 L.

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Melted wax has a temperature of 54 deg. Celsius. Physical or chemical property?
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The melting of solid wax to form liquid wax and the evaporation of liquid wax to form wax vapor are physical changes. The burning of the wax vapor is a chemical change.

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What volume of a 6.67 M NaCl solution contains 3.12 mol NaCl
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