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3 years ago

Which of these elements is a metal? A. hafnium (Hf) B. radon (Rn) C. silicon (Si) D. sulfur (S) E. selenium (Se)

Chemistry

2 answers:

Semenov [28]3 years ago

4 0

Answer: Option (A) is the correct answer.

Explanation:

Substances or elements which lose electrons are known as metals whereas elements which gain electrons are known as non-metals.

For example, electronic configuration of hafnium is $[Xe]4f^{14}5d^{2}6s^{2}$ . It is a transition metal as it has incompletely filled d-orbital so, it will lose electrons to attain stability.

Electronic configuration of radon is $[Xe]4f^{14}5d^{10}6s^{2}6p^{6}$ . Since it is a noble gas it has completely filled orbitals so, it will neither lose nor gain electrons.

Electronic configuration of silicon is $[Ne]3s^{2}3p^{2}$ and it is a non-metal. So, it will gain electrons to complete its octet.

Electronic configuration of sulfur is $[Ne]3s^{2}3p^{4}$ and it is a non-metal. So, it will gain electrons to complete its octet.

Electronic configuration of selenium is $[Ne]3d^{10}4s^{2}4p^{4}$ and it is a non-metals. So, it will gain electrons to complete its octet.

Hence, we can conclude that out of the given options hafnium (Hf) is the element which is a metal.

Len [333]3 years ago

3 0

The answer is A. Hafnium

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Explanation:

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3 years ago

What is the mass of oxygen in 3.34 g of potassium permanganate?

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Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

Molar Mass of KMnO₄ = 158 g/mol

Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).

Mass contributed by Oxygen (O) = 4 (16) = 64 g

Since the percentage of compound is 100

So,

Percent of Oxygen (O) = 64 / 158 x 100

Percent of Oxygen (O) = 40.5 %

It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.

So,

for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

mass of Oxygen (O) = 0.405 x 3.34 g

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4 years ago

Explain how the following experimental errors affect the final calculation of the kilocalories per gram for a food item. Be spec

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Answer:

a) the final kilocalories per gram for food will be less because the mass was reduced

b)the final kilocalories per gram for food will be less since

c) the final kilocalories per gram for food will be less since the reaction will eventually go to completion

d) the final kilocalories per gram for food will be more.

Explanation:

a) the final kilocalories per gram for food will be less because the mass was reduced from 110.3 to 101.3g

b)the final kilocalories per gram for food will be less since some marshmallow fell off before the reaction

c) the final kilocalories per gram for food will be less since the reaction will eventually go to completion

d) the final kilocalories per gram for food will be more since the thermometer that got stuck will add to the value of final kilocalories per gram

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4 years ago

2.6(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64°C. The standard enthalpy of vapori

Sophie [7]

Answer:

The value of W is 5.602 kJ, Q is -70.6 kJ, change in U is -65 kJ, and change in H is -70.3 kJ.

Explanation:

Based on the given information, the mass of CH3OH given is 64 grams, which is condensed isothermally and reversibly to liquid at 64 degrees C. The given standard enthalpy of vaporization of methanol at 64 degrees C is 35.3 kJ per mole.

The moles of CH3OH can be determined by using the formula,

Moles = Mass / Molar mass

= 64.0 grams / 32.0 grams per mole

= 2 mol

The amount of energy given by the process of condensation is,

ΔH = 2 mol × 35.3 kJ/mol = 70.6 kJ

In condensation heat is given off, thus, it is an exothermic process, hence, q will be -70.6 kJ

The work or W can be calculated by using the formula,

W = -P ΔV

Let us first find the volume of 2.0 mole gas at 64 °C, or 64 + 273 = 337 K,

PV = nRT

V = nRT/P

= 2 mol × 0.08206 L atm per mol K × 337 K/1 atm

= 55.3 L

As the liquid condenses in the process, the change in volume would be negligible. So, the volume change will be -55.3 L

W = - 1 atm × - 55.3 L

W = 55.3 L.atm

W = 55.3 L.atm × 101.3 J/1 L atm = 5602 J

W = 5602 × 1 kJ / 1000 J = 5.602 kJ

W = 5.602 kJ

Now U can be calculated using the formula,

U = q + W

= -70.6 kJ + 5.602 kJ

= -65. kJ

Thus, q = -70.6 kJ, W = 5.602 kJ, U = -65 kJ, and ΔH = -70.3 kJ.

4 0

3 years ago