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3 years ago

Explain why the atomic radius decreases across a period and increases down a group ?

Chemistry

1 answer:

andreyandreev [35.5K]3 years ago

4 0

Answer:

In general, atomic radius decreases across a period and increases down a group. ... Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.

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In Experiment 9, a 1.05 g sample of a mixture of NaCl (MW 58.45) and NaNO2 (MW 69.01) is reacted with excess sulfamic acid. The

Kaylis [27]

Answer:

There is 76.6 mL of nitrogen collected

Explanation:

Step 1: Data given

Mass of the sample = 1. 05 grams

The sample is 40.00% by mass NaNO2

MW of NaCl = 58.45 g/mol

MW of NaNO2 = 69.01 g/mol

Temperature = 22.0 °C

Pressure = 750.0 mmHg = (750/760) atm

The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm

Step 2: Calculate mass of NaNO2

(40/100)*1.05 = 0.42 grams

Step 3: Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

Step 4: Calculate moles of N2

For 2 moles of NaNO2 we'll get 1 mol of N2

For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

Step 5: Calculate pressure of N2

P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa

Step 6: Calculate volume of N2

PV = nRT

⇒ P = the pressure of N2 = 0.96079

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = moles of N2 = 0.00304 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C = 295 Kelvin

V = (0.00304*0.08206*295)/0.96079

V = 0.0766 L = 76.6 mL

There is 76.6 mL of nitrogen collected

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4 years ago

Uranium-238 decays very slowly.its half-life is 4.47 billion years.

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13.4 billion years is 3 times of the half-life, 4.47 billion years. So the Uranium-238 will go through three times of half decay. So the remain percentage will be 50%*50%*50%=12.5%.

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3 years ago

Read 2 more answers

Uranium-235 and uranium-238 are considered isotopes of one another

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If you are asking which is the most abundant, Uranium-238 is

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4 years ago

A 0.08024 M solution of NaOH was used to titrate a solution of unknown concentration of HCl. A 32.08 mL sample of the HCl soluti

iris [78.8K]

Answer:

0.08024

Explanation:

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3 years ago

The base-dissociation constant, kb, for pyridine, c5h5n, is 1.4x10-9 the acid-dissociation constant, ka, for the pyridinium ion,

sp2606 [1]

We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:

Kw = Kb * Ka

where Kw is always equal to 1x10^-14

so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:

1x10^-14 = 1.4x10^-9 * Ka
solve for Ka

Ka = 7.14x10^-6

Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.

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3 years ago

Explain why the atomic radius decreases across a period and increases down a group ?​

Explain why the atomic radius decreases across a period and increases down a group ?