How is scientific use of the term digital different from the common use?

Final answer:

Propane's complete combustion reaction with oxygen producing carbon dioxide and water is a chemical process, representing a typical exothermic combustion reaction used for heat generation.

Explanation:

The combustion of propane (C3H8) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water vapor (H2O) is a chemical reaction that involves the breaking and forming of chemical bonds.

The balanced chemical equation for this reaction is:

The balanced chemical equation for the complete combustion of propane is C3H8 + 5O2 → 3CO2 + 4H2O. This reaction is exothermic, releasing heat energy which is why propane is commonly used as a fuel source.

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.

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.

The new volume of the gas can be calculated using Boyle's law of gases. The volume of the gas when pressure drops from 90 kpa to 20 kpa, increases from 4 L to 18 L.

Boyle's law states that, at constant temperature, the volume of a gas is inversely proportional to the pressure of the gas.

hence, PV = a constant.

Let P1 and V1 be the initial pressure and volume and P2 and V2 be the final pressure and volume, then

P1 V1 = P2V2.

Given that,

P1 = 90 kPa

P2 = 20 kPa

V1 = 4 L

V2 = P1 V1/P2

     = (90kPa  × 4 L)/20 kPa

     = 18 L.

Therefore, the final volume of the gas will be 18 L.

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Answer: Option (a) is the correct answer.

Explanation:

A substance that does not get consumed in a chemical reaction and helps in increasing rate of the reaction by lowering the activation energy is known as a catalyst.

Hence, when the chemist finds that reaction speeds up when platinum is added into the reaction then the chemist must calculate or measure the mass of platinum before and after the completion of reaction.

This is because if there occurs change in the mass of platinum then it means that platinum is not acting as a catalyst in the reaction.

Thus, we can conclude that chemist must measure the mass of the platinum to determine whether it is a catalyst.

This is a true statement.

Core science provides a basic understanding of key scientific concepts, while double science goes further in depth and grants two GCSE qualifications upon completion. The choice between the two often depends on a student's academic and career aspirations in the field of science.

The main difference between core science and double science as part of the England GCSE OCR exam board involves the extent of the curriculum covered and the number of GCSE qualifications granted upon completion.

Core Science is a single award course which delivers one GCSE qualification. This course provides a broad and basic understanding of key scientific concepts, covering aspects of Biology, Chemistry, and Physics.

On the other hand, Double Science, also known as Combined Science, awards two GCSE qualifications. It covers a more extensive curriculum, diving deeper into the subjects of Biology, Chemistry, and Physics. This might be the preferred choice for students wishing to pursue sciences at A-level or beyond, as it provides a more comprehensive foundation in these subject areas.

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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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Answer:

7.56 moles

Explanation:

For a 742 kPa, we can consider the gas as an ideal gas because it will be close to 7 atm (1 kPa = 0.01 atm), which is low pressure. So, the ideal gas equation is:

PV = nRT

Where P is the pressure, V is the volume, n the number of moles, R the gas constant and T the temperature (always in Kelvin). For pressure in kPa and volume in liters, R = 8.314 kPa.L/mol.K, then:

742x25 = nx8.314x295

2452.63n = 18550

n = 18550/2452.63

n = 7.56 moles

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.

"the addition of negatively charged electrons" is the answer.

[tex]\boxed{{\text{7}}{\text{.182 g}}}[/tex] of [tex]{\text{CuS}}{{\text{O}}_{\text{4}}}[/tex] would be needed to make 150 mL of 0.300 M [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution.

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is as follows:

[tex]{\text{Molarity of CuS}}{{\text{O}}_4}\;{\text{solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}}}{{{\text{Volume }}\left( {\text{L}}\right){\text{ of}}\;{\text{CuS}}{{\text{O}}_4}\;{\text{solution}}}}[/tex]   ......(1)

Rearrange equation (1) to calculate the moles of [tex]{\text{CuS}}{{\text{O}}_4}[/tex].

[tex]{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}= \left( {{\text{Molarity of HN}}{{\text{O}}_3}\;{\text{solution}}} \right)\left( {{\text{Volume of}}\;{\text{CuS}}{{\text{O}}_4}\;{\text{solution}}} \right)[/tex]   ......(2)

The volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is to be converted into L. The conversion factor for this is,

[tex]{\text{1 mL}} = {10^{ - 3}}\;{\text{L}}[/tex]

So the volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is calculated as follows:

[tex]\begin{aligned}{\text{Volume of CuS}}{{\text{O}}_4}\;{\text{solution}}&= \left( {{\text{150 mL}}}\right)\left({\frac{{{{10}^{ - 3}}\;{\text{L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.15{\text{ L}}\\\end{aligned}[/tex]

The molarity of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is 0.300 M.

The volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is 0.15 L.

Substitute these values in equation (2).

[tex]\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}&=\left( {{\text{0}}{\text{.300 M}}} \right)\left( {0.15{\text{ L}}}\right)\\&= 0.04{\text{5 mol}}\\\end{aligned}[/tex]

The formula to calculate the mass of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] is as follows:

[tex]{\text{Mass of CuS}}{{\text{O}}_{\text{4}}}=\left( {{\text{Moles of CuS}}{{\text{O}}_{\text{4}}}} \right)\left( {{\text{Molar mass of CuS}}{{\text{O}}_{\text{4}}}} \right)[/tex]       ......(3)

Substitute 0.045 mol for the moles of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] and 159.609 g/mol for the molar mass of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] in equation (3).

[tex]\begin{aligned}{\text{Mass of CuS}}{{\text{O}}_{\text{4}}}&= \left({{\text{0}}{\text{.045 mol}}} \right)\left( {\frac{{{\text{159}}{\text{.609 g}}}}{{{\text{1 mol}}}}}\right)\\&= 7.1824{\text{05 g}}\\&\approx {\text{7}}{\text{.182 g}}\\\end{aligned}[/tex]

Therefore the mass of [tex]{\mathbf{CuS}}{{\mathbf{O}}_{\mathbf{4}}}[/tex]  is 7.182 g.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, CuSO4, dilution, moles of CuSO4, volume, solution, 0.045 mol, 150 mL, 0.300 M, 7.182 g, 159.609 g/mol, 0.15 L.

B. This looks like ice turning to water then to steam

Answer is: B.) A solid gains kinetic energy to become a liquid and then becomes a gas.

Stage 2 is melting (from solid to liquid) and Stage 2 is boiling (from liquid to gas).

In solid, molecules are closely packed, stiff and do not changes of shape or volume. Solid object does not take on the shape of its container.

Liquids have definite volume, but no fixed shape.

Gases not have definite volume and fixed shape, it depends on its container.

Answer: The correct answer would 100% be A.  Experiments showed that the atom is made of smaller particles.

just took the test and got it right

The number of molecules in  12.5 g water is equal to 4.18 × 10²³ molecules.

Avogadro’s constant can be defined as the proportionality constant that uses to count the number of constituents such as ions, molecules, and atoms in the amount of substance in a sample.

Avogadro's constant can be defined as the approximate number of nucleons per gram of substance. The value of the Avogadro number is the mass of 1 mole of a compound, in grams, and is equal to the number of nucleons in 1 constituent particle.

The value of Avogadro’s constant is experimentally found to be approximately equal to 6.022×10²³ per mole.

The mass of the water = 12.5 g

The molar mass of the water = 18 g

The number of moles of the water = 12.5/18 = 0.694 mol

The number of molecules of water in one mole = 6.022 × 10²³

Number of molecules of water in 0.694 mol = 6.022 × 10²³× 0.694

= 4.18 × 10²³ molecules

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