In the context of a balanced chemical equation, a____is a number to the lower right of an element or ion within formula and it applies to____. A____is a number to the left of a formula and it applies to_____. Only_____should be added or changed when balancing chemical equations.

Answers

Answer 1
Answer:

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.

Answer 2
Answer:

Final answer:

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.

Explanation:

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.

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The image formed by a lens may be real or virtual. The image formed by a lens is always virtual.

Answers

There are types of lens; concave and convex lens.
The concave lens is a lens which has an inward curve in the middle, that is, the edges of the curve are thicker than the center of the lens, because of this, any light that enter the lens will spread out [diverge]. An image will look smaller and upright when viewed by a concave lens. Image formed by concave lens are usually VIRTUAL. 
A concave lens will produce a real image ONLY if the object is located beyond the focal point of the lens.
A convex lens is a converging lens, this is because, the center of the lens is thicker than its edges. Any ray of light that passes through the lens will converge at the middle of the lens at point called principal focus.  A convex lens produce a VIRTUAL image when the object is placed infront of the focal point. The virtual image formed is always magnified and upright.

The image formed by any lens like concave and convex may be real or virtual.

How many types of lens are present?

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.

  • Any light passing through the lens will disperse [diverge]. When viewed through a concave lens, an image seems smaller and more upright. Concave lenses frequently produce VIRTUAL images and concave lens will produce a real image only if the object is located beyond the focal point of the lens.
  • Any beam of light passing through the lens will converge in the lens's center, known as the main focus. When an object is placed in front of the focal point, a convex lens produces a VIRTUAL image. The resulting virtual picture is constantly amplified and oriented vertically.

Hence image formed by a lens may be real or virtual.

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Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.

Answers

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]

pH = 7.1581

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-cells. The reduction potential of Ni2+ is �0.23 V. Calculate the potential of the cell at 25�C if the more dilute Ni2+ solution is in the anode compartment.

Answers

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  Ni(s)\rightarrow Ni^(2+)+2e^-

Reduction half reaction (cathode):  Ni^(2+)+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_(cell) will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Ni^(2+)_(diluted)])/([Ni^(2+)_(concentrated)])

where,

n = number of electrons in oxidation-reduction reaction = 2

E_(cell) = ?

[Ni^(2+)_(diluted)] = 1.00* 10^(-4)M

[Ni^(2+)_(concentrated)] = 1.0 M

Putting values in above equation, we get:

E_(cell)=0-(0.0592)/(2)\log (1.00* 10^(-4)M)/(1.0M)

E_(cell)=0.118V

Hence, the cell potential of the cell is +0.118 V

The  cell potential for the cell as calculated  is 0.118 V.

What is the Nernst equation?

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.

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If a molecule diffusing through extracellular fluid travels 1 mm in 1 sec, how long will it take that molecule to diffuse 1 cm

Answers

given that the rate of diffusion is 1 mm per 1 second. then the time it travels in 1 cm can be solve using the formula
t = d / r
where d is the distance
r is the rate

first, 1 cm is equal to 10 mm

t = 10 mm / ( 1 mm / s )
t = 10 s

Of the following elements, which one has the lowest first ionization energy?boron carbon aluminum silicon

Answers

Answer:

Boron

Explanation:

Because it has a complete 2s orbital and therefore, an increased shielding of the 2s orbital will reduce the ionisation energy.

Final answer:

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.

Explanation:

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.

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A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1860 N to the right. The forces acting on the crate are

Answers

are radioactive waves , hope this helps