Which of these is a major source of organic compounds? petroleum coal natural gas all of these

We commonly know refer to something 'digital' has to something electronic that can be visibly seen such as a watch, clock, camera, screen, etc.  It really refers to stored energy or electricity that's not natural.  But the word 'digital' in science refers to the depiction of data or information in figures (such as in a table) in contrast to as a chart, graph, drawing, or other pictorial form.

Explanation:

The scientific use of the term digital is much changed from the current use because the term digital in science is a way of representing a discontinuous phenomenon or something this is just finite and faulty. It also describes a data in figures with the use of this term. While “digital” generally refers to electronics in general, the scientific explanation of digital is much different

The table of values for four aqueous solutions at   is as shown in Table 1.

Further Explanation:

The negative logarithm of hydrogen ion concentration is termed as pH while that of hydroxide ion concentration is pOH. Higher the pH of solution, more will be the alkalinity and vice-versa. Similarly, higher value of pOH indicates solution is acidic in nature .

The expression for pH is mentioned below.

[tex]\text{pH}=-\text{log}\left[\text{H}^+\right][/tex]                                             ...... (1)

Where [tex]\left[\text{H}^+\right][/tex]  is the concentration of hydrogen ion.

The expression for pOH is mentioned below.

[tex]\text{pOH}=-\text{log}\left[\text{OH}^-\right][/tex]                                        ...... (2)

Where [tex]\left[\text{OH}^-\right][/tex]  is the concentration of hydroxide ion.

pH and pOH are related to each other by following expression:

pH + pOH = 14                                               …… (3)

Solution A:

Substitute [tex]6.4\times10^{-13}\text{ M}[/tex]  for [tex]\left[\text{H}^+\right][/tex]  in equation (1).

[tex]\begin{aligned}\text{pH}=&-\text{log}\left[6.4\times10^{-13}\text{ M}\right]\\=&12.2\end{aligned}[/tex]

Rearrange equation (3) for pOH.

pOH = 14 – pH                                                                          …… (4)

Substitute 12.2 for pH in equation (4).

[tex]\begin{aligned}\text{pOH}=&14-12.2\\=&1.8\end{aligned}[/tex]

Rearrange equation (2) for [tex]\left[\text{OH}^-\right][/tex]  .

[tex]\left[\text{OH}^-\right]=10^{-\text{pOH}}[/tex]                                                                        …… (5)

Substitute 1.8 for pOH in equation (5).

[tex]\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{1.8}}\\=&0.016\text{ M}\end{aligned}[/tex]

Solution B:

Substitute  [tex]2.7\times10^{-10}\text{ M}[/tex] for [tex]\left[\text{OH}^-\right][/tex]  in equation (2).

[tex]\begin{aligned}\text{pOH}=&-\text{log}\left[2.7\times10^{-10}\text{ M}\right]\\=&\ 9.6\end{aligned}[/tex]

Rearrange equation (3) for pH.

pH = 14 – pOH                                                                                 …… (6)

Substitute 9.6 for pOH in equation (6).

[tex]\begin{aligned}\text{pH}=&14-9.6\\=&4.4\end{aligned}[/tex]

Rearrange equation (1) for [tex]\left[\text{H}^+\right][/tex] .

[tex]\left[\text{H}^+\right]=10^\text{-pH}[/tex]                                                                                        …… (7)  

Substitute 4.4 for pH in equation (7).

[tex]\begin{aligned}\left[\text{H}^+\right]=&10^\text{-4.4}\\=&0.0000398\text{ M}\end{aligned}[/tex]

Solution C:

Substitute 8.11 for pH in equation (7).

[tex]\begin{aligned}\left[\text{H}^+\right]=&10^\text{-8.11}\\=&0.0000000078\text{ M}\end{aligned}[/tex]

Substitute 8.11 for pH in equation (4).

[tex]\begin{aligned}\text{pOH}=&14-8.11\\=&5.89\end{aligned}[/tex]

Substitute 5.89 for pOH in equation (5).

[tex]\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{5.89}}\\=&0.000001288\text{ M}\end{aligned}[/tex]

Solution D:

Substitute 4.73 for pOH in equation (5).

[tex]\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{4.73}}\\=&0.00001862\text{ M}\end{aligned}[/tex]

Substitute 4.73 for pOH in equation (6).

[tex]\begin{aligned}\text{pH}=&14-4.73\\=&9.27\end{aligned}[/tex]

Substitute 9.27 for pH in equation (7).

[tex]\begin{aligned}\left[\text{H}^+\right]=&10^\text{-9.27}\\=&0.0000000005\text{ M}\end{aligned}[/tex]

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Answer details:

Grade: High School

Chapter: Acids, base and salts

Subject: Chemistry

Keywords: pH, pOH, H+, OH-, solution A, solution B, solution C, solution D, hydrogen ion, hydroxide ion, negative logarithm.

Answer:

c

Explanation:

3.38x10^-3G

Answer:

D. separates the lower three-quarters of the data set from the upper quarter

Explanation:

The upper quartile is a line or a point that divides the upper half of a set of data in two exact parts, so when you get the median of the whole set of data, you take the upper half and calculate the median of that upper half, dividing the upper half in quartiles, and you get the division between the upper quartile and the other three quartiles.

Ammonia acts as a Brønsted-Lowry base because it accepts a hydrogen ion from water, forming ammonium and hydroxide ions. It also satisfies the Arrhenius definition by increasing the concentration of hydroxide ions in the solution. Thus, the answer most appropriately corresponds to option b, Brønsted-Lowry definition.

When ammonia, [tex]NH_3[/tex], dissolves in water, it functions as a base according to the Brønsted-Lowry definition because it accepts a hydrogen ion (H+) from water to form ammonium ions ([tex]NH_4^+[/tex]) and hydroxide ions ([tex]OH^-[/tex]). Water acts as a Brønsted-Lowry acid in this reaction because it donates a hydrogen ion to the ammonia. This process forms a conjugate acid-base pair, with ammonia being the base that becomes the conjugate acid ([tex]NH_4^+[/tex]), and water being the acid that becomes the conjugate base ([tex]OH^-[/tex]). Additionally, ammonia also meets the Arrhenius definition of a base because the reaction results in an increase in the concentration of hydroxide ions in the solution. Thus, the answer most appropriately corresponds to option b, Brønsted-Lowry definition.

In the compound Al2(CrO4)3, the aluminum ion (Al) has a 3+ charge. To balance the electrical neutrality, each chromate ion (CrO4) must have a 2- charge since three of them are needed to balance the 6+ charge from two aluminum ions.

Chromate ion has a charge of 2-. When forming an ionic bond with aluminum ions which have a 3+ charge, the formula becomes Al2(CrO4)3.

To determine the ion charge for the chromate ion in the compound Al2(CrO4)3, we must first understand the charges of the ions involved. Aluminum (Al) forms a cation with a 3+ charge, as indicated by the formula. Since the formula of the compound is electrically neutral and contains two aluminum ions, the total positive charge contributed by aluminum is 6+ (2 ions * 3+).

The compound then must have an equivalent total negative charge to balance this, contributed by three chromate (CrO4) ions. For the charges to balance out, each chromate ion must carry a 2- charge (6+ from Al3+ is balanced by 6- from three 2- ions), making the formula for the ionic compound Al2(CrO4)3 electrically neutral. Therefore, the chromate ion has a 2- charge in this compound.

Answer: when the mixture is homogeneous

Explanation:

Answer : The number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

Explanation :

As we know that the 1 mole of substance contains [tex]6.022\times 10^{23}[/tex] number of atoms.

As per question,

The 1 mole sample of neon contains [tex]6.022\times 10^{23}[/tex] number of atoms.

And,

As, 1 mole sample of argon contains [tex]6.022\times 10^{23}[/tex] number of atoms

So, 4.0 mole sample of argon contains [tex]4.0\times 6.022\times 10^{23}=24.088\times 10^{23}[/tex] number of atoms

From this we conclude that, the number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

The particles are moving fastest in Region D) 5.

As temperature increases, so does the average kinetic energy (speed) of the molecules.

Answer: The correct option is (D)

Explanation:

As the temperature increases the energy of the particles increases which results in increase of kinetic energy of the particles. With this increase in energy particles moves more rapidly.

So, the particles at higher temperature will be moving at faster speed in region 5.

To find out how many moles of lead iodide are produced from 246.2 g of potassium iodide, first convert the mass of KI to moles using its molar mass, then divide by the stoichiometric ratio from the reaction. It is found that 246.2 g of KI will produce 0.7415 moles of lead iodide.

To determine how many moles of lead iodide are produced from 246.2 g of potassium iodide (KI), it is necessary to consider the stoichiometry of the reaction given by:

[tex]Pb(NO_3)_2 (aq) + 2 KI (aq) \rightarrow PbI_2 (s) + 2 KNO_{3} (aq)[/tex]

First, we calculate the number of moles of KI using its molar mass:

1 mole KI = 166.0028 g/mol. Therefore,

Moles of KI = 246.2 g

166.0028 g/mol = 1.483 moles of KI

Since the reaction requires 2 moles of KI for every mole of [tex]PbI_2[/tex], we divide the moles of KI by 2 to find the moles of [tex]PbI_2[/tex]:

[tex]PbI_2[/tex]moles = 1.483 moles / 2 = 0.7415 moles of [tex]PbI_2[/tex]

Therefore, 246.2 g of KI will produce 0.7415 moles of lead iodide.

Answer:

0.092 M

Explanation:

NaOH(aq) + HCl(aq) ------> NaCl(aq) + H2O(l)

CA= concentration of acid= 0.1000 M

CB= concentration of base= ?????

VA= volume of acid= 25.00 mL

VB= volume of base= 27.16 mL

NA= number of moles of acid= 1

NB= number of moles of base= 1

CAVA/CBVB= nA/nB

CB= CAVAnB/VBnA

CB= 0.1000 M × 25.00 mL ×1/27.16 mL ×1

CB= 0.092 M

Answer:2.23 moles

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass , occupies 22.4 Liters at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given volume}}{\text{Molar volume}}=\frac{50}{22.4}=2.23moles[/tex]

Thus 2.23 moles of ammonia gas are required to fill a volume of 50 liters at STP.

This is a question based on colligative property

colligative property is a property of solution which depends upon the number of molecules of solute and solvent

It is independent of nature of solute added to a particular solvent

The colligative properties are

a) elevation in boiling point

b) depression in freezing point

c) relative lowering of vapour pressure

d) osmotic pressure

As given we are adding same mount of each substance

Which means that we are adding same moles of each substance

the minimum number of particles will be produced by CH3OH (methanol) hence it will cause least temperature change.

The greatest change will be due to K3PO4

The modern periodic table is largely organized by the atomic number of each element. Although atomic weight and chemical properties also play a role, they come secondary to atomic number.

The modern periodic table is primarily organized on the basis of the atomic number of each element. The atomic number is equal to the number of protons in the nucleus of an atom of an element. While other factors such as atomic weight and chemical properties are also significant, they are considered secondary in the structure. For example, Hydrogen has an atomic number of 1, hence it is the first element in the Periodic Table. Similarly, Helium with an atomic number of 2 is the second element and so on.

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