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blsea [12.9K]
2 years ago
15

How does the atomic radius change as you move left to right in Period 3 from aluminum through chlorine? Refer to the periodic ta

ble.
A. it generally increases.

B. it does not change.

C. it generally decreases.

D. it varies unpredictably.

Chemistry
1 answer:
Serggg [28]2 years ago
7 0

Answer:

C it generally decrease

Explanation:

Atomic radius of an element decrease from left to right of the periodic table

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A 0.708 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4
ollegr [7]

Answer:

The metal has a molar mass of 65.37 g/mol

Explanation:

Step 1: Data given

Mass of the metal = 0.708 grams

Volume of hydrogen = 275 mL = 0.275 L

Atmospheric pressure = 1.0079 bar = 0.9947 atm

Temperature = 25°C

Vapor pressure of water at 25 °C = 0.03167 bar = 0.03126 atm

Step 2: The balanced equation

M(s) + H2SO4(aq) ⟶ MSO4 (aq) + H2(g)

Step 3: Calculate pH2

Atmospheric pressure = vapor pressure of water + pressure of H2

0.9947 atm = 0.03126 atm + pressure of H2

Pressure of H2 = 0.9947 - 0.03126

Pressure of H2 = 0.96344 atm

Step 4: Calculate moles of H2

p*V=n*R*T

⇒ with p = The pressure of H2 = 0.96344 atm

⇒ with V = the volume of H2 = 0.275 L

⇒ with n = the number of moles H2 = TO BE DETERMINED

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

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.96344 * 0.275)/(0.08206*298)

n = 0.01083 moles

Step 5: Calculate moles of M

For 1 mole of H2 produced, we need 1 mole M

For 0.0108 moles of H2 we need 0.01083 moles of M

Step 6: Calculate molar mass of M

Molar mass M = Mass M / moles M

Molar mass M = 0.708 grams / 0.01083 moles

Molar mass M = 65.37 g/mol

The metal has a molar mass of 65.37 g/mol

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Explanation:

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<span>Answer: Glass has a much higher specific heat than styrofoam so more of the energy produced will go into heating the glass. This quantity of heat will be lost (unaccounted for) hence the delta t in the calculation will be too small and the heat of neutralization will be lower.</span>
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