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kondor19780726 [428]
3 years ago
11

Using the measurements in the table, determine which unidentified metal has the lowest density

Chemistry
2 answers:
Oliga [24]3 years ago
3 0
C is the correct answer because density is mass per unit volume and when you divide it we that volume is greater than mass so density is less density density is directly proportional to Mass Density is inversely proportional to volume
xxTIMURxx [149]3 years ago
3 0

Answer:

\boxed{\text{Metal c}} }

Explanation:

\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\textbf{a. }\text{Density} = \dfrac{\text{122 g}}{\text{12.5 cm}^{3}} = \text{9.96 g/cm}}^{3}\\\\\textbf{b. }\text{Density} = \dfrac{\text{132 g}}{\text{14.2 cm}^{3}}= \text{9.30 g/cm}}^{3}\\\\\textbf{c. }\text{Density} = \dfrac{\text{126 g}}{\text{18.1 cm}^{3}}= \text{6.96 g/cm}}^{3}\\\\\textbf{d. }\text{Density} = \dfrac{\text{126 g}}{\text{12.7 cm}^{3}}= \text{9.92 g/cm}^{3}\\\\\boxed{\textbf{Metal c}}\text{ has the lowest density}

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Explain why the ionization energy to remove a second electron from potassium is higher than the ionization energy to remove four
never [62]

Explanation:

It is more difficult to remove electrons from the second shell or energy level because of the imbalance between the positive nuclear charge and the remaining electrons.

  • The amount of energy required to remove electrons in ground state of an atom is the ionization energy.
  • The first ionization energy is the energy needed to remove the most loosely bound electron of an atom in the gas phase in ground state.
  • The second energy has a greater nuclear pull as it is closer to the nucleus.
  • Both potassium and silicon have the same number of energy levels.
6 0
3 years ago
What happens when an electron absorbs energy?
OLga [1]
The correct answer is A I believe
4 0
3 years ago
how many liters of a 60% antifreeze solution must be added to 8L of a 10% antifreeze solution to produce a 20% antifreeze soluti
Tasya [4]

Answer: 2Liters

Explanation:

The expression used will be :

M_1V_1+M_2V_2=M_3V_3

where,

M_1 = concentration of first antifreeze= 60%

M_2 = concentration of second  antifreeze= 10%

V_1 = volume of first antifreeze = x L

V_2 = volume of second antifreeze = 8 L

M_3 = concentration of final antifreeze solution= 20%

V_3 = volume of final antifreeze = (x+8) L

Now put all the given values in the above law, we get the volume of  antifreeze added

60\times x+10\times 8=20\times (x+8)

x=2L

Therefore, the volume of 60% antifreeze solution that must be added is 2L

5 0
3 years ago
What is the natural rate of nitrogen fixation in Earth’s ecosystems? What is the natural rate of nitrogen fixation in Earth’s ec
quester [9]

Answer:

100 teragrams of nitrogen per year

Explanation:

Nitrogen fixation in Earth's ecosystems is defined as a process where by nitrogen in air is transformed into ammonia or other related nitrogenous compounds. Generally, atmospheric nitrogen is referred to as molecular dinitrogen and it is a nonreactive compound that is metabolically useless to all but a few microorganisms. This process is vital to life due to the fact that inorganic nitrogen compounds are needed for the biosynthesis of amino acids, protein, and all other nitrogen-containing organic compounds. Thus, the natural rate of nitrogen fixation in Earth's ecosystems is 100 tetragrams of nitrogen per year.

7 0
3 years ago
Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
3 years ago
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