Answer:
Lead (II) Sulfate
Explanation:
Don't forget to add the (II) marking indicating that lead has a charge of 2+ (because Sulfate has a charge of 2-, balancing the ion).
Answer:
B. Mixing a solute and a solvent
Explanation:
Hello,
In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.
Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.
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It looks like we are solving for a pressure. All that is required is some algebraic manipulation to find our pressure in mmHg.
Given:
(5.0 m³)(7.5 mmHg) = (P)(4.0m³)
Multiply:
37.5 = 4.0P
Divide:
9.375 = P
P = 9.4 mmHg (remember sig figs)
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9.4 mmHg
Answer:
Resonance structures are represented in the picture below.
Explanation:
When there is a double pair of electrons shared between atoms in a molecule, the position of these electrons can be changed, without changing the molecule conformation. This occurs to stabilization, the electrons are relocated. These structures are called resonance structures.
In the molecule of NO₂F, nitrogen has 5 electrons in its valence shell, so it needs 3 electrons to be stable. Oxygen has 6 electrons and needs 2 to be stable, and fluor has 7 electrons and needs one electron to be stable.
Nitrogen still has electrons after the sharing, so it can also share one pair and will have a partial positive charge. One of the oxygens will not complete the octet, so will share only one pair f electron and will have a partial negative charge, that will compensate the positive charge in nitrogen.
The two resonance structures are shown below:
Answer:
153.6771 amu
Explanation:
From the question given above, the following data were:
Isotope A:
Mass of A = 114.3789 amu
Abundance (A%) = 64.23%
Isotope B:
Mass of B =.?
Abundance (B%) = 100 – A%
Abundance (B%) = 100 – 64.23
Abundance (B%) = 35.77%
average atomic mass of Element Y = 128.4359 amu
The mass of the 2nd isotope (i.e isotope B) can be obtained as follow:
Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
128.4359 = [(114.3789 × 64.23)/100] + [(Mass of B × 35.77) /100]
128.4359 = 73.4656 + (Mass of B × 0.3577)
Collect like terms
128.4359 – 73.4656 = Mass of B × 0.3577
54.9703 = Mass of B × 0.3577
Divide both side by 0.3577
Mass of B = 54.9703 / 0.3577
Mass of B = 153.6771 amu
Therefore, the mass of the 2nd isotope is 153.6771 amu