You want to go swimming, and the pH of a water in your swimming pool is 7.9. You want it to be between 7.4-7.6. What should you add to the pool to lower its pH?

1. Water
2. Alkali
3. Acid

In the electronic circuit, the function of battery is to push the electrons through the circuit.

An electronic circuit comprises of individual electronic components like resistors, diodes,transistors , capacitors,etc. which are connected to each other by means of wires through which electric current flows.It can also be referred to as an electronic circuit.

It usually comprises of at least one active component . The combination of wires and simple and complex components allows operations to be performed, amplification of signals, displacement of data and performing of computations.

There are three types of circuits , an analog circuit ,a digital circuit and a mixed circuit which is combination of digital and analog circuits , even semiconductors are used in circuits.

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Answer:

Mass = 8.46 g

Explanation:

Given data Mass of water produced = ? Mass of glucose = 20 g Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂     →   6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

              C₆H₁₂O₆           :            H₂O

                  1                   :              6

                0.11                :           6/1×0.11 = 0.66

                  O₂               :            H₂O

                  6                   :              6

                0.47                :           0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol  ×18 g/mol

Mass = 8.46 g

Explanation:

Answer:

An ester of sulfuric acid in which only one sulfuric acid hydroxyl group has been esterified. The product of esterification of both sulfuric acid hydroxyl groups is a sulfate ester.

The reactions that result in the emission of light involve the ruthenium label and tripropylamine (TPA), two electrochemically active molecules.

Thus, The electrode surface inside the measurement cell is where the reactions take place.

The ruthenium label is oxidized at the electrode surface as an electrical potential is applied, and TPA is oxidized into a radical cation that spontaneously loses a proton.

When the resultant TPA radical interacts with oxidized ruthenium, the ruthenium label enters an excited state and emits a photon (620 nm) before decaying. The ruthenium label is renewed and ready to carry out numerous light-generating cycles as it goes back to its ground state.

Thus, The reactions that result in the emission of light involve the ruthenium label and tripropylamine (TPA), two electrochemically active molecules.

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Answer:

Examples of ROS include peroxides, superoxide, hydroxyl radical, singlet oxygen, and alpha-oxygen.

In the given reaction, the limiting reactant is C₆H₆ (benzene).

To determine the limiting reactant as well as calculate the mass of HCl produced, compare the moles of each reactant.

The number of moles for each reactant:

Molar mass of Cl₂ = 35.5 g/mol + 35.5 g/mol = 71 g/mol

Moles of Cl₂ = mass of Cl₂ / molar mass of Cl₂

                     = 45 g / 71 g/mol

                     = 0.634 moles of Cl₂

Molar mass of C₆H₆ (benzene) = 12 g/mol + 6(1 g/mol) = 78 g/mol

Moles of C₆H₆ = mass of C₆H₆ / molar mass of C₆H₆

                        = 45 g / 78 g/mol = 0.577 moles of C₆H₆

Determine the stoichiometry between Cl₂ and HCl:

Cl₂ + C₆H₆ → C₆H₅Cl + HCl

Here, we can see that 1 mole of Cl₂ produces 1 mole of HCl.

Thus, the limiting reactant is C₆H₆ (benzene).

Calculate the mass of HCl produced:

Molar mass of HCl = 1 g/mol + 35.5 g/mol = 36.5 g/mol

Moles of HCl produced = moles of C₆H₆ = 0.577 moles

Mass of HCl produced = moles of HCl produced × molar mass of HCl

Mass of HCl produced = 0.577 moles × 36.5 g/mol

                                     ≈ 21.04 g

Therefore, approximately 21.04 grams of HCl will be produced.

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Answer:

Choose three different light levels, and place four identical plants under each light level to observe the light’s effect on multiple plants.

Explanation:

Answer:Prokaryotic cells carry out cellular respiration within the cytoplasm and eukaryotic cells carry out respiration in the mitochondria

Explanation:

Answer:

First is battery acid with a pH of 1, second is stomach acid with a pH of 1.5-3.5 (depends), third is antacid with a pH of 7, fourth is baking soda with a pH of 9, and finally, fifth is household ammonia with a pH of 11.

Here is a picture of the pH levels to better help you understand.

Answer:

b

Explanation:

A)  the addition of \(H_3O\)+ or OH- ions would have opposing effects on the buffer solution.

b) \(H_3O\)+ would favor the formation of acetic acid, while adding OH- would favor the formation of acetate ions.

c) The exact magnitude of the changes in concentrations depends on the initial concentrations of \(CH_3COOH\) and \(CH_3COONa\), as well as the specific amount of \(H_3O\)+ or OH- added.

To determine the effect of adding \(H_3O\)+ and OH- on the given buffer solution, we need to consider the ionization of acetic acid (\(CH_3COOH\)) and the dissociation of its sodium salt (CH3COONa). Let's analyze each scenario separately:

A) Buffer solution consisting of 0.5 M\(CH_3COOH\) and 0.5 M \(CH_3COONa\):

When acetic acid (CH3COOH) and its sodium salt (\(CH_3COONa\)) are present together in a solution, they form a buffer system. Acetic acid partially ionizes in water, releasing \(H_3O\)+ ions, while sodium acetate dissociates into Na+ and \(CH_3COO\)- ions.

Adding \(H_3O\)+:

The \(H_3O\)+ ions would react with the acetate ions (CH3COO-) to form undissociated acetic acid (\(CH_3COOH\)) through the following reaction:

\(H_3O\)+ + \(CH_3COO\)- ⇌ \(CH_3COOH\) + H2O

The addition of H3O+ would shift the equilibrium to the left, promoting the formation of more acetic acid and decreasing the concentration of acetate ions.

Adding OH-:

The OH- ions would react with the acetic acid (\(CH_3COOH\) to form water and acetate ions (CH3COO-) through the following reaction:

OH- + \(CH_3COOH\) ⇌ \(CH_3COO\)- + H2O

The addition of OH- would shift the equilibrium to the right, consuming acetic acid and increasing the concentration of acetate ions.

B) After adding 0.02 mol of NaOH to 1.0 L of the buffer solution:

When solid NaOH is added to the buffer solution, it dissociates completely in water to form Na+ and OH- ions.

NaOH dissociation:

NaOH → Na+ + OH-

The OH- ions formed would react with acetic acid according to the reaction mentioned in Scenario A (2), increasing the concentration of acetate ions and consuming acetic acid.

C) After adding 0.02 mol of HCl to 1.0 L of the buffer solution:

When HCl is added to the buffer solution, it dissociates completely in water to form \(H_3O\)+ and Cl- ions.

HCl dissociation:

HCl → \(H_3O\)+ + Cl-

The \(H_3O\)+ ions formed would react with acetate ions (\(CH_3COO\)-) according to the reaction mentioned in Scenario A (1), forming more undissociated acetic acid and decreasing the concentration of acetate ions.

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Answer:

63.33%

Explanation:

To find the percentage of sugar in the milk chocolate just divide 15 by 9.5. This will give you 0.63333 (recurring). To find the percentage, times it by 100, which will give you 63.33%.

Hope this helped :)

The percentage of sugar present in the milk would be  63.33 %.

In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.

As given in the problem a 15.00 g milk chocolate bar contains 9.500 grams of sugar, and we have calculated the percentage of sugar present,

The percentage of sugar present in the milk = 9.500 ×100/15

                                                                                = 63.33 %

Thus, the percentage of sugar present in the milk would be 63.33 %.

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Each sentence about the bioavailability of minerals in foods should be completed with either increases or decreases as follows:

Nutrients can be defined as the chemical compounds (substances) or minerals that are present in food, and they are needed by the body of a living organism for healthy growth, reproduction, development, proper functioning, and reproduction.

In Science, there are five (5) main classes of nutrients and these include the following:

In conclusion, phytic acid, oxalic acid, and refinement of grains are factors that decrease the bioavailability of minerals in foods while Vitamin C and Vitamin D increases the bioavailability of minerals in foods.

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To graph the function g(x) = f(-x), you can start with the graph of f(x) and then reflect it about the y-axis.

To find the graph of the function g(x) = f(-x), we can start with the graph of the function f(x) and then reflect it about the y-axis.

If the graph of f(x) is symmetric with respect to the y-axis, meaning it is unchanged when reflected, then g(x) = f(-x) will have the same graph as f(x).

However, if the graph of f(x) is not symmetric with respect to the y-axis, then g(x) = f(-x) will be a reflection of f(x) about the y-axis.

In either case, the resulting graph of g(x) = f(-x) will be symmetric with respect to the y-axis.

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Answer:

Concentration: The amount of a given substance within a set solution.

Explanation:

For example, the units for concentration include mol/\(dm^{3}\) which means that if a substance has 3 mol/\(dm^{3}\), it means 3 moles of that substance is in a 1 dm cubed of that solution. You could give examples of the concentration of acids like Hydrochloric Acid (HCl) and alkalis like Hydroxide (OH).

The chemical equation for the decomposition of water is:

\(2 H_2O -- > 2 H_2 + O_2\)

To balance this equation, we need to count the number of atoms of each element on both sides of the equation.

On the left side of the equation, we have:

2 hydrogen atoms (2 H₂O)

2 oxygen atoms (2 H₂O)

On the right side of the equation, we have:

2 hydrogen atoms (2 H₂)

2 oxygen atoms (1 O₂)

We can see that the number of hydrogen atoms is already balanced, but the number of oxygen atoms is not. To balance the equation, we need to add a coefficient in front of O2 so that we have the same number of oxygen atoms on both sides.

The balanced equation is:

\(2 H_2O -- > 2 H_2 + 1 O_2\)

A compound is broken down into simpler compounds during a decomposition reaction. Different techniques, such as heating, exposure to light, or the inclusion of a catalyst, can be used to produce this reaction.

The reactant component splits into two or more products, which may be elements or compounds, during decomposition. A synthesis reaction, in which less complex substances join to create a more complex compound, is the antithesis of this reaction.

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Answer:

b

Explanation:

b

"Yes, a precipitate will occur since Ksp is less than Q," is the response.

Based on the given information, we can calculate the reaction quotient (Q) for the precipitation of barium carbonate:

BaCl2(aq) + Na2CO3(aq) → BaCO3(s) + 2NaCl(aq)

Q = [Ba2+][CO32-] = (1.5 x 10^-3 M)(1.0 x 10^-3 M) = 1.5 x 10^-6

Since the value of Q (1.5 x 10^-6) is greater than the value of the solubility product constant (Ksp = 2.6 x 10^-9), the reaction is not at equilibrium and will proceed to form a precipitate of barium carbonate (BaCO3).

Therefore, the answer is "Yes, a precipitate will form because Ksp is less than Q".

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KCl = Potassium chloride

NaCl = Sodium chloride

FeCl2 = Iron (II) chloride or ferrous chloride

MgO = Magnesium oxide

CuO = Copper (II) oxide or cupric oxide

LiCl = Lithium chloride

PbO = Lead (II) oxide or plumbous oxide

CuI = Copper (I) iodide or cuprous iodide

NaOCl = Sodium hypochlorite

SF4 = Sulfur tetrafluoride

Hope this helped !

the pressure required for 1.35 g of oxygen to dissolve in 1.5 L of water is 3.56 atm.

The first step in solving this problem is to identify the relevant equation.

Henry's law is the formula that relates the pressure of a gas above a liquid to the concentration of the gas that dissolves in the liquid.

In mathematical terms, Henry's law can be expressed as follows:P = kH * Cwhere P is the pressure of the gas, kH is Henry's law constant, and C is the concentration of the gas in the liquid.

To solve the problem, we need to first determine the value of kH using the given data.

kH can be calculated using the following formula:kH = P / CSubstituting the values given in the problem into this formula, we get:kH = 1.65 atm / (0.654 g / 1.5 L) = 3.97 atm/(g/L).

Now that we have the value of kH, we can use Henry's law to calculate the pressure required for 1.35 g of oxygen to dissolve in 1.5 L of water.

To do this, we simply rearrange the formula to solve for P:P = kH * CSubstituting the values of kH and C into this formula, we get:P = 3.97 atm/(g/L) * (1.35 g / 1.5 L) = 3.56 atm

Therefore, the pressure required for 1.35 g of oxygen to dissolve in 1.5 L of water is 3.56 atm.

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Answer:

8 electrons

Explanation:

Number of atoms = 128

Half life = 15 s

After each half life, the number of atoms reduces by half. So we need to determine how many half lives are in 60s.

Number of half lives = 60s / 15s = 4

After the first life;

Number of electrons left = 128 / 2 = 64

After the second life;

Number of electrons left = 64 / 2 = 32

After the third half life;

Number of electrons left = 32 / 2 = 16

After the fourth half life;

Number of electrons left = 16 / 2 = 8

Answer:

C. 8

Explanation:

correct on gizmo; After one half-life (15 seconds), about 64 radioactive atoms will remain, half of the original 128. After a second half-life passes (30 seconds total), about 32 radioactive atoms will remain, half of 64. After a third half-life passes (45 seconds), about 16 radioactive atoms will remain. A time of 60 seconds represents four half-lives, so about 8 radioactive atoms will remain.

The fact that the infrared thermometers used in medical research were first intended to detect the temperature on Mars has had an impact on that field of study.

Technology for use in outer space, in aviation or other operations outside of Earth's atmosphere, for things like spaceflight, space exploration, and Earth observation, is referred to as space technology.

By enabling high-speed data transport with the development of the internet, enabling research of global natural phenomena, and enabling environmental monitoring, space technology started to be employed for telecommunications and the environment.

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Answer:

\(\huge\boxed{\sf M = 0.06\ M}\)

Explanation:

No. of moles = n = 0.03 mol

Volume = v = 500 ml = 0.5 L

Molarity = M = ?

M = n / v

Put the given data in the above formula.

M = 0.03 / 0.5

\(\rule[225]{225}{2}\)

Answer:c

Explanation:c

Answer:

Explanation:

we know that

ΔH=m C ΔT

where ΔH is the change in enthalpy (j)

m is the mass of the given substance which is water in this case

ΔT IS the change in temperature and c is the specific heat constant  

we know that given mass=2.9 g

ΔT=T2-T1 =98.9 °C-23.9°C=75°C

specific heat constant for water is 4.18 j/g°C

therefore ΔH=2.9 g*4.18 j/g°C*75°C

ΔH=909.15 j

The correct option is (d) To express very large or very small number.

What is the Scientific Notation?

Scientific notation is a way to present numbers that are too large or too small to be easily written in decimal form. The three components of  scientific notation are coefficient, base and exponent. The proper format to write a scientific notation is a x 10^b, where a is a number or decimal number and b is the power of 10 to make scientific notation equivalent to original number. When a number between 1 and 10 is multiplied by a power of 10 then the number is expressed in scientific notation. For example,  10000000 can be written as 10⁷, which is the scientific notation and the exponent is positive here. Similarly, for the negative exponent 0.000001 can be can be represented as 10-⁷.                             Hence, the scientific notation is used to express very large or very small numbers.

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If an object is changing directions, it can be moving at a constant speed while yet accelerating.

This is due to the fact that acceleration is a vector number that takes into consideration the speed of an object both in magnitude and direction.

An object's acceleration vector also changes as it changes direction, which causes the object's velocity vector to change.

As a result, even though the item is moving at a constant speed, the change in its velocity vector is causing it to accelerate.

Centripetal acceleration is the term for the type of acceleration that keeps objects moving in a circular motion.

Even though an object's speed is constant while it moves in a circular pattern, it is still accelerating.

This is due to the fact that acceleration is a vector number that takes into consideration the speed of an object both in magnitude and direction.

As an item follows a circular course, its acceleration vector also constantly changes direction along with its velocity vector.

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While tumor marker test results can be useful, they are not conclusive. A low result does not imply that you do not have cancer or that you are in remission.

CA 15-3, for example, is raised in less than half of patients with early breast cancer and in more than 80% of those with metastatic breast cancer. Three tumor markers, cancer antigen 15-3 (CA 15-3), cancer antigen 27.29 (CA 27.29), and carcinoembryonic antigen (CEA), have been used in breast cancer care to help monitor metastatic breast cancer (advanced disease), but they have not been found to be useful in detecting a breast cancer recurrence or extending lives. If the level falls, the treatment is effective. If it rises, the cancer may be spreading.

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Answer:

39.119 g

Explanation:

1 mole of S has 32.065 g

=> 1.22 x 32.065 = 39.1193 or 39.119 g