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3 years ago

What is the mass of a 12.31 cm3 piece of copper having a density of 8.96 g/cm3?

Chemistry

1 answer:

malfutka [58]3 years ago

7 0

Answer:

mass is 110.3g

Explanation:

Density = mass/volume

density = 8.96 g/cm³

volume = 12.31cm³

mass = ?

density x volume = mass

8.96 x 12.31 = 110.2976g≈110.3g

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During inhalation, View Available Hint(s) During inhalation, oxygen molecules move into the lungs, and carbon dioxide molecules

bekas [8.4K]

Answer:

the diaphragm and rib muscles contract

Explanation:

Inhalation is the set of movements that allows air to enter the lungs. In this process occurs the contraction of the diaphragm, causing its lowering. The intercostal muscles also contract causing the ribs to lift. This causes the chest to grow larger and the internal pressure of the lungs to become smaller than the external. With the reduction of intrapulmonary pressure, air ends up entering the airways and reaching the alveoli for gas exchange to occur.

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4 years ago

Jack observed that the growth of certain plants in his orchard is not proper. He thinks that it is because of some hormonal prob

natta225 [31]

Answer:

abscisic acid

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3 years ago

How many grams of Ca are contained in 113 mL of 0.52 M Ca (OH)2 solution

Kay [80]

113 ml = 0.113L

0.52M = 0.52 mol/L

(0.52 mol/L) x 0.113L = 0.059 mol x (molar mass of Ca) = your answer.

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3 years ago

For the following pair, indicate which element has the lower first ionization energy: Match the words in the left column to the

emmainna [20.7K]

Answer:

For 1: The correct answer is Ge.

For 2: The correct answer is Te.

For 3: The correct answer is Ba.

For 4: The correct answer is Ag.

Explanation:

Ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. It is represented as $E_i$

$X(g)\rightarrow X^+(g)+1e^-;E_i$

Ionization energy increases as we move from left to right in a period. This happens because the atomic radius of an element decreases moving across a period, which increases the effective attraction between the negatively charged electrons and positively-charged nucleus. Hence, the removal of electron from the outermost shell becomes difficult and requires more energy.

Ionization energy decreases on moving from top to bottom in a group. This happens because the number of shells increases as we move down the group. The electrons get added in the new shell. So, the shielding of outermost electrons from the inner ones is more which decreases the attraction between the electrons and the nucleus. Hence, the removal of electron from the outermost shell becomes easy and requires less energy.

For the given options:

Option 1:

Chlorine is the 17th element of the periodic table belonging to Period 3 and Group 17.

Germanium is the 32nd element of the periodic table belonging to Period 4 and Group 14.

Hence, germanium will have smaller first ionization energy.

Option 2:

Tellurium is the 52nd element of the periodic table belonging to Period 5 and Group 16.

Selenium is the 34th element of the periodic table belonging to Period 4 and Group 16.

Hence, tellurium will have smaller first ionization energy.

Option 3:

Barium is the 56th element of the periodic table belonging to Period 6 and Group 2.

Titanium is the 22nd element of the periodic table belonging to Period 4 and Group 4.

Hence, barium will have smaller first ionization energy.

Option 4:

Copper is the 29th element of the periodic table belonging to Period 4 and Group 11.

Silver is the 47th element of the periodic table belonging to Period 5 and Group 11.

Hence, silver will have smaller first ionization energy.

5 0

3 years ago

Given the enthalpies of reaction

Tju [1.3M]

2SO₂+O₂⇒2SO₃ ΔH=-197 kJ

<h3>Further explanation</h3>

Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways

Reaction

2SO₂+O₂⇒2SO₃

Given :

1. S(s)+O₂(g)→SO₂(g) ΔH = -297 kJ

Reverse

SO₂(g) ⇒S(s)+O₂(g ΔH = +297 kJ (sign change to +) x 2

2SO₂(g) ⇒2S(s)+2O₂(g ΔH = +594 kJ

2.2S(s)+3O2(g)→2SO3(g) ΔH=-791kJ

Add both reactions and remove/subtract the same compound for different sides

1. 2SO₂(g) ⇒2S(s)+2O₂(g) ΔH = +594 kJ

2.2S(s)+3O₂(g)→2SO₃(g) ΔH=-791kJ

--------------------------------------------------------+

2SO₂+O₂⇒2SO₃ ΔH=-197 kJ

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3 years ago