Answer:
Explanation:
The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH4) and two diatomic oxygen molecules (O2) to produce one carbon dioxide molecule (CO2) and two water molecules (H2O). The chemical equation representing this process is provided in the upper half of Figure 1, with space-filling molecular models shown in the lower half of the figure.
In a balanced chemical equation, a subscript is a number to the right of an element indicating the number of atoms in a molecule. A coefficient is a number to the left of a formula indicating the number of molecules. Only coefficients should be altered when balancing equations.
In the context of a balanced chemical equation, a subscript is a number to the lower right of an element or ion within a formula and it applies to the number of atoms of that element in a molecule. A coefficient is a number placed to the left of a formula and it applies to the number of molecules of the entire substance. Only coefficients should be changed when balancing chemical equations because altering subscripts changes the substance itself.
The balanced chemical equation is a symbolic representation of a chemical reaction, where the number of each type of atom is equalized for both the products and reactants, in accordance with the law of conservation of matter.
For example, in the equation 2H₂O, the subscript '2' to the right of 'H' shows that there are two hydrogen atoms in one water molecule, and the coefficient '2' to the left of 'H₂O' means there are two molecules of water, totaling four hydrogen atoms.
#SPJ2
The image formed by any lens like concave and convex may be real or virtual.
Mainly two types of lens are present and that are concave lens also known as divergent lens and convex lens, also known as convergent lens.
Hence image formed by a lens may be real or virtual.
To know more about concave & convex lens, visit the below link:
#SPJ3
Answer:
pH = 7.1581
Explanation:
The equilibrium of NaHSO₃ with Na₂SO₃ is:
HSO₃⁻ ⇄ SO₃²⁻ + H⁺
Where K of equilibrium is the Ka2: 6.5x10⁻⁸
HSO₃⁺ reacts with NaOH, thus:
HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺
As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:
HSO₃⁻: 0.252 moles
SO₃²⁻: 0.139 moles
Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:
0.0500L ₓ (1mol /L): 0.050 moles of NaOH.
Thus, final moles of both compounds are:
HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles
SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles
Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:
pH = pka + log [SO₃²⁻] / [HSO₃⁻]
Where pKa is - log Ka = 7.187
Replacing:
pH = 7.187 + log [0.189] / [0.202]
Answer: The cell potential of the cell is +0.118 V
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode):
Reduction half reaction (cathode):
In this case, the cathode and anode both are same. So, will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
=
= 1.0 M
Putting values in above equation, we get:
Hence, the cell potential of the cell is +0.118 V
The cell potential for the cell as calculated is 0.118 V.
The Nernst equation can be used to obtain the cell potential of a cell under non- standard conditions. The standard cell potential in this case is zero owing to the fact that both cathode and anode are made of nickel.
Hence;
Ecell = E°cell - 0.0592/nlog Q
Ecell = 0 - 0.0592/2 log (1 00 * 10^-4/1)
Ecell = 0.118 V
The cell potential for the cell as calculated is 0.118 V.
Learn more about Nernst equation: brainly.com/question/721749
Answer:
Boron
Explanation:
Because it has a complete 2s orbital and therefore, an increased shielding of the 2s orbital will reduce the ionisation energy.
Among boron, carbon, aluminum, and silicon, aluminum has the lowest first ionization energy due to its position on the periodic table, which is further to the left and in a higher period than the other elements.
Ionization energy refers to the energy necessary to remove an electron from an atom in its gaseous state. The element with the lowest first ionization energy among boron, carbon, aluminum, and silicon is aluminum. Ionization energy increases from left to right across a period in the periodic table and from bottom to top in a group. Thus, aluminum, being to the left of boron, carbon, and silicon, has the lowest first ionization energy. Furthermore, aluminum is in the third period, which is below boron and carbon's second period.
#SPJ3