A 1.00 x 10^-10 M solution of chloric acid Please help me solve this to find pH and pOH​

Answer:

i'm pretty sure its the last answer and if not its the 2nd one

Answer:

1. lakes and rivers 2. groundwater 3. watershed 4. pollution

Explanation:

Answer:

See explanation.

Explanation:

Hello there!

In this case, for the required chemical reactions, we proceed as follows, by considering that the left-handed elements stand for the cations and the right-handed ones for the anions:

1:

1. [tex]Na_2CO_3(aq)\rightarrow 2Na^+(aq)+(CO_3)^{2-}(aq)[/tex]

2. [tex]FeBr_3(aq)\rightarrow Fe^{3+}(aq)+3Br^-(aq)[/tex]

3. [tex]Al_2(SO_4)_3(aq)\rightarrow 2Al^{3+}(aq)+3(SO_4)^{2-}(aq)[/tex]

2: In this case, we can represent the formula equation first according to the neutralization reaction:

[tex]3NaOH(aq)+H_3PO_4(aq)\rightarrow Na_3PO_4(aq)+3H_2O(l)[/tex]

Which is worded as: three molecules of aqueous sodium hydroxide reacts with one molecule of aqueous phosphoric acid to yield one molecule of aqueous sodium chloride and three molecules of liquid water.

Regards!

Answer:

Number of Protons 48

Number of Neutrons 64

Number of Electrons 48

Explanation:

The mass number of an atom is the sum of the number of protons and neutrons in the nucleus of that atom

Answer:

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Answer:

C) pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

Explanation:

To find the pH of a buffer (The mixture of the weak acid and its conjugate base) we have to use Henderson-Hasselbalch equation:

pH = pKa + log [Base] / [Acid]

Where pKa is the pka of the buffer and [] is molar concentration of the species of the buffer

The [Base] is equal to the concentration of NaOH added:

0.10M * (0.005L / 0.015L)

And the concentration of the acid [Acid] is the initial concentration of the acid - the concentration of the NaOH added:

0.0100L * (0.20M)/0.0150L - 0.10M * (0.005L / 0.015L)

That means the pH of the buffer is:

Answer:

C) pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

Explanation:

The equilibrium for the acid ionization of HC2HO3 is represented by the equation above. If 10.0mL of 0.20MHC2HO3 reacts with 5.0mL of 0.10MNaOH, which of the following could be used to calculate the correct pH of the resulting solution. pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

Answer:

Bacteria are also involved in many processes that are indispensable for our life on Earth. Many of these processes have to do with the recycling (reuse) of chemical elements that have been here since the formation of the planet

Explanation:

I hope it will help you

Domains eukarya  would this organism most likely fit.

The domain eukarya comprised of eukaryotes or organisms whose cells contain true nucleus.

It is  the largest of all groups in the classification of life. There are three domains:-

It is the domain of organism called eukaryotes. These are the organism who have a well defined nucleus and membrane bound organelles.

Hence, C) option is correct.

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Answer:  A

The third law is the law of Newton’s Law of motion.

Answer: The diagonal relationship is present between the lighter elements of periods second, thirds.

Explanation:

Elements which are present diagonally adjacent to each other in the second and third periods of the periodic table tends to show diagonal relationship as they have similar properties.

For example, Lithium is a second period element and magnesium is a third period element.

Hence, lithium and magnesium show diagonal relationship with each other.

Thus, we can conclude that diagonal relationship is present between the lighter elements of periods second, thirds.

Answer:

0.322 g

Explanation:

Since our concentration is given in mass per volume percent % (m/v)

% (m/v) = mass of solute in g/volume of solution in mL × 100%

Since % (m/v) = 0.92 and volume of the required solution is 35.0 mL, we find the mass of NaCl from

mass of NaCl = % (m/v)/100 % × volume of solution

= 0.92 % (m/v)/100 %× 35 mL

= 0.0092 × 35

= 0.322 g

The study of a chemical is called chemistry.

The correct answer is 0.322 g

The saline is said that the amount of salt present in the solution. The formula used to solve the question is as follows:-

[tex]\frac{m}{v} = \frac{mass \ of\ solute}{volume\ of\ solution} * 100[/tex]

The data is given as follows:-

Mass of NaCl is as follows:-

 [tex]\frac{0.92}{100} * 35 mL[/tex]

[tex]0.0092 * 35\\= 0.322 g[/tex]

Hence, the correct answer is 0.322g.

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Answer:

Option A

Explanation:

Addition of too much of solvent will make the solution dilute due to which the crystals will not form. Hence option D is incorrect

On the other hand adding a minimum amount of boiling solvent will give a saturated solution for recrystallization. Hence, option A is incorrect

Addition of cold solvent will lower the rate of formation of crystals. Hence, both option B and C are incorrect

Explanation:

sexual reproduction has minimal changes of mutations compared to asexual reproduction

Answer:

sexual reproduction has minimal changes of mutations compared to asexual reproduction

I hope this helps

By using combined gas law , volume will be 1.88 [tex]dm^{3}[/tex].

The combined gas law contains the laws of Charles, Boyle, and Gay-Lussac. When you know the pressure, volume, and temperature of a system and need to modify one of these, you can apply this law to compute the resulting unknown.

Combined gas law is expressed as:

[tex]$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$[/tex]

Calculation of volume by using combined gas law

It is given that, [tex]V_{1}[/tex] = 254[tex]dm^{3}[/tex], at STP P = 101.3 kPa and T = 273 K

Now, put the value of given data in above equation.

[tex]V_{2}[/tex] = (254[tex]dm^{3}[/tex]) (101.3 kPa) (26+273) /(150 kPa) (273)

[tex]V_{2}[/tex] = 1.88 [tex]dm^{3}[/tex].

By using combined gas law  volume will be 1.88 [tex]dm^{3}[/tex].

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Answer:

Warm front

Explanation:

A warm front forms when a warm air mass pushes into a cooler air mass, shown in the image to the right (A). Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms. Warm fronts move more slowly than cold fronts because it is more difficult for the warm air to push the cold, dense air across the Earth's surface. Warm fronts often form on the east side of low-pressure systems where warmer air from the south is pushed north.

You will often see high clouds like cirrus, cirrostratus, and middle clouds like altostratus ahead of a warm front. These clouds form in the warm air that is high above the cool air. As the front passes over an area, the clouds become lower, and rain is likely. There can be thunderstorms around the warm front if the air is unstable.

On weather maps, the surface location of a warm front is represented by a solid red line with red, filled-in semicircles along it, like in the map on the right (B). The semicircles indicate the direction that the front is moving. They are on the side of the line where the front is moving. Notice on the map that temperatures at ground level are cooler in front of the front than behind it.

Answer:

Normality factor is defined as the ratio of observed wt. of solute to the theoretical wt. of the solute required to prepare a solution of desired normality. ... The principle states that, “the product of volume and concentration factor of the solution at any dilution is constant provided mass of the solution is constant.”

Explanation:

Answer:

The answer is B

Explanation:

The answer is B because representative particles can only be atoms.

Answer:

c

Explanation:

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Answer:

The entropy increases

Explanation:

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Answer:

See Explanation

Explanation:

Metallic bonds involve attraction between electrons and positively charged metal ions. The metals are ionized and electrons form a sea of valence electrons. These loosely bound electrons surround the nuclei of the metals.

The presence of this sea of electrons explains the fact that metals conduct electricity and heat due to the free valence electrons.

Due to the nature of the bonding between metal atoms,metals are malleable and ductile.

Due to the strong electrostatic interaction between metal ions and electrons, the metallic bond is very strong and is very difficult to break thereby accounting for the greater strength of metals as the size of the metallic ion decreases.

Answer:

In the most simplest of words: The number of atoms in a molecule

ex) c6h12o6

There are 6 carbon atoms and oxygen atoms, 12 hydrogen atoms

Another example NH4

there are 1 nitrogen atoms and 4 hydrogen atom

Hope that answers your question

Explanation:

Answer:

ch3

Explanation:

Answer:

CERES is the latgest in the solar system

Explanation:

Answer:

OKAY EVERYONE STOP SAYING PANCREAS THATS WRONG

Explanation:

It is actually thyroid

Answer:

Volume is reduced to half

Explanation:

Acid base titration are commonly used reactions in a lab, and are ofter used to get pH or different kind of solutions.

The neutralization of an acid base reaction is reached, when the solution (having added an indicator previously) changes its original color. chemically speaking, this occurs when the number of moles of the acid and the base are balanced and equal. In other words the following:

n₁ = n₂   (1)

This expression can also be expressed in function of concentration and volume:

M₁V₁ = M₂V₂   (2)

From here, solving for V₁:

V₁ = M₂V₂  / M₁

Now, this expression is true only when we have the same kind of substance that can lose or gain the same number of hydrogens.

Lets suppose that we have as base NaOH (Monoprotic base) and HCl (monoprotic acid), the titration reaction would be:

NaOH + HCl --------> NaCl + H₂O

As both of the species are monoprotic, the number of moles are the same  when they reach the equilibrium, so, expression  (2) can be used, and calculate volume or concentrations.

However, in this case, a partner made a mistake and use a diprotic acid, in this case, H₂SO₄, In this case, things chance because H₂SO₄ is diprotic, meaning that we need to dissociate two hydrogens in equilibrium, therefore, expression (2) would be something like this.

Acid: 1; Base: 2

H₂SO₄ + 2NaOH ------> Na₂SO₄ + H₂O

nH₂SO₄ = n₁ = 1

nNaOH = n₂ = 2

n₁/n₂ = 1/2

2n₁ = n₂   (3)

Writting this, in function of concentration and volume, it would be:

2M₁V₁ = M₂V₂   (4)

From here, if we solve for the volume of the acid (V₁):

V₁ = M₂V₂ / 2M₁

Therefore, according to this expression, we can see that the volume required of the acid would be half the volume required of the monoprotic acid. For example, if we need 50 mL of Chloridic acid to reach the equivalence point with NaOH, then, with H₂SO₄ it will only need 25 mL. This, of course, assuming that concentrations are the same, and volume of the base used, the same.

Hope this helps

Answer:

i dont knoiwxplanation:

Answer:

1. Volume of stock required = 15 mL

Volume of water required = 485 mL

2. Volume of stock required = 5 mL

Volume of water required = 495 mL

3. Volume of stock required = 5.30 mL

Volume of water required = 494.70 mL

4. Volume of stock required = 50 mL

Volume of water required = 450 mL

Explanation:

Using the dilution formula C₁V₁ = C₂V₂

Where C₁ = initial concentration; C₂ = final concentration; V₁ = initial volume; V₂ = final volume

1. C₁ = 5 M NaCl, C₂ = 150 mM NaCl = 0.15 M NaCl

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 0.15 × 500/ 5 = 15 mL

Volume of stock required = 15 mL

Volume of water required = 500 mL - 15 mL

Volume of water required = 485 mL

2. C₁ = 1 M Tris-HCl, C₂ = 50 mM Tris-HCl = 0.05 M Tris-HCl

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 0.05 × 500/ 5 = 5 mL

Volume of stock required = 5 mL

Volume of water required = 500 mL - 5 mL

Volume of water required = 495 mL

3. C₁ = 95% NP-40, C₂ = 1% NP-40

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 1 × 500/ 95 = 5.30 mL

Volume of stock required = 5.30 mL

Volume of water required = 500 mL - 5.30 mL

Volume of water required = 494.70 mL

4. C₁ = 1% SDS, C₂ = 10% SDS

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 1 × 500/ 10 = 50 mL

Volume of stock required = 50 mL

Volume of water required = 500 mL - 50 mL

Volume of water required = 450 mL