Metal D Most reactive
Sodium
Magnesium
Carbon
Metal E
Iron
Hydrogen
Copper Least reactive

Answer:  There are

Explanation:

1) Diesel has a higher viscosity than petrol.

2) Petrol is more flammable than diesel.

3) The formula will be C₁₀H₂₂.

4) The equation is; 2C8H18+25O2→16CO2+18H2O.

Depending on the precise composition and temperature, the viscosity of gasoline and diesel can change. In general, diesel is more viscous than gasoline. Higher viscosity fluids are thicker and flow more slowly than lower viscosity fluids because viscosity relates to the resistance of a fluid to flow.

Diesel is less flammable than gasoline. The lowest temperature at which gasoline can evaporate and turn into an ignitable combination in air is known as its flash point, and it is lower for gasoline. Compared to diesel fuel, petrol vapors are much more flammable and can ignite at lower temperatures.

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The word equations tells us that the combination of copper oxide and sulfuric acid produces copper sulfate and water

A word equation is a way to represent a chemical reaction using words instead of chemical formulas or symbols. It describes the reactants and products of the reaction in a clear and understandable manner as we see in the question that was shown as equation.

Note that the copper oxide and the sulfuric acids are the reactants that are combined and the copper sulfate and the water are the products.

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Explanation:

An oxidizer is a substance that facilitates oxidation, a chemical reaction where a substance loses electrons. It is also called an oxidizing agent or oxidant. Oxidizers are commonly used in combustion reactions, supporting the burning of fuels by providing oxygen. Examples of oxidizers include oxygen, chlorine, hydrogen peroxide, and potassium permanganate. They have applications in combustion, chemical synthesis, bleaching, rocket propellants, and cleaning. Oxidizers can be highly reactive and require proper handling and safety precautions.

Answer: Therefore, the concentration of hydroxide ions [OH-] in the given solution is 4.0 x 10⁻¹⁹M.

We know that In an aqueous solution Kw is the ionization constant of water.

Kw = [H3O⁺][OH⁻]

[OH⁻] = [tex]\frac{Kw}{[H3O^+]}[/tex]--------------------------------------(a)

Kw = ionization constant of water

[H3O⁺]= the concentration of hydronium ions

[OH⁻]  =  the concentration of hydroxide ions

Kw = 1x10⁻¹⁴M²-------------------(i)

[H3O⁺]= 2.5 x 10⁴M------------------(ii)

[OH⁻]  = ?    

NOW Putting values in  (i) and (ii) in equation (a)

[OH⁻] = [tex]\frac{1 X 10^-^1^4}{2.5 X 10^4}[/tex]

[OH⁻] =  4.0 x 10⁻¹⁹M

As per the given data, 1.6 grams of oxygen reacted in this experiment.

To calculate the amount of oxygen that reacted in the experiment, we need to determine the difference in the mass of X oxide (X,O) formed and the mass of X initially used.

Given:

Mass of X = 4.6 g

Mass of X oxide (X,O) = 6.2 g

To find the amount of oxygen that reacted:

Mass of oxygen = Mass of X oxide - Mass of X

= 6.2 g - 4.6 g

= 1.6 g

Therefore, 1.6 grams of oxygen reacted in this experiment.

Calculate the mass of 1 mole of X:

Given that the mass of X is 4.6 g, we can calculate the molar mass of X by dividing the mass by the number of moles:

Molar mass of X = Mass of X / Number of moles of X

Molar mass of X = 4.6 g / 0.1 mol

Molar mass of X = 46 g/mol

Therefore, the mass of 1 mole of X is 46 grams.

Thus, the answer is 46 grams.

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A) The chemical equation for the reaction is [tex]Cl_2(aq) + 2KBr(aq)[/tex] ⟶ [tex]2KCl(aq) + Br_2(aq)[/tex]

B)No reaction occurs when elemental iodine is mixed with lithium chloride. [tex]I_2(aq) + 2LiCl(aq)[/tex]⟶ No Reaction

A) To predict whether a reaction will occur when elemental chlorine ([tex]Cl_2[/tex]) and potassium bromide (KBr) are mixed, we can refer to the activity series for the halogens. According to the activity series, chlorine is more reactive than bromine. Therefore, chlorine can displace bromine from its compounds.

The chemical equation for the reaction between chlorine and potassium bromide can be written as:

[tex]Cl_2(aq) + 2KBr(aq)[/tex] ⟶ [tex]2KCl(aq) + Br_2(aq)[/tex]

In this reaction, chlorine displaces bromine from potassium bromide, resulting in the formation of potassium chloride and elemental bromine.

B) Similarly, to predict whether a reaction will occur when elemental iodine ([tex]l_2[/tex]) and lithium chloride (LiCl) are mixed, we can refer to the activity series. In the halogen activity series, iodine is less reactive than chlorine and bromine. Therefore, it is less likely for iodine to displace chlorine or bromine from their compounds.

The chemical equation for the reaction between iodine and lithium chloride can be written as:

[tex]I_2(aq) + 2LiCl(aq)[/tex]⟶ No Reaction

No reaction occurs because iodine is less reactive than chlorine, and lithium chloride does not react with iodine under these conditions.

Therefore, when elemental chlorine is mixed with potassium bromide, a reaction occurs and chlorine displaces bromine. On the other hand, no reaction occurs when elemental iodine is mixed with lithium chloride.

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The molecular formula of the compound is P₂O₅.

To determine the molecular formula for a compound with the given percent compositions, we can assume a certain mass for the compound and calculate the number of moles for each element based on its percentage. Then, we can determine the simplest whole-number ratio of the elements to find the molecular formula.

Let's assume a mass of 100 grams for the compound. From the given percent compositions, we have 43.7 grams of phosphorus (P) and 56.3 grams of oxygen (O).

Next, we calculate the moles of each element using their molar masses. The molar mass of phosphorus (P) is 30.97 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.

The moles of phosphorus can be calculated as:

moles of P = mass of P / molar mass of P = 43.7 g / 30.97 g/mol ≈ 1.41 mol

The moles of oxygen can be calculated as:

moles of O = mass of O / molar mass of O = 56.3 g / 16.00 g/mol ≈ 3.52 mol

Now, we find the simplest whole-number ratio of the elements by dividing the moles of each element by the smallest number of moles (1.41 mol in this case):

P:O ≈ 1.41 mol / 1.41 mol : 3.52 mol / 1.41 mol ≈ 1:2.5

Since we need whole-number ratios, we can multiply both numbers by 2 to get a whole number for oxygen:

P:O ≈ 2:5

Therefore, the molecular formula of the compound is P₂O₅.

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The concentration of hydroxide ions ([OH-]) and the concentration of hydronium ions ([H+]) are related in an aqueous solution by the equation [H+][OH-] = 1.0 x 10^-14 at 25 °C .The concentration of hydronium ions ([H+]) in the aqueous solution at 25 °C is approximately 3.03 x 10^-18.

Given that [OH-] is 3.3 x 10^3, we can substitute this value into the equation as follows:

[H+][3.3 x 10^3] = 1.0 x 10^-14

Dividing both sides of the equation by 3.3 x 10^3, we get:

[H+] = (1.0 x 10^-14) / (3.3 x 10^3)

Simplifying the expression, we have:

[H+] ≈ 3.03 x 10^-18

In summary, at 25 °C, an aqueous solution with an OH- concentration of 3.3 x 10^3 has a hydronium ion concentration of approximately 3.03 x 10^-18. The hydronium ion concentration is determined by the equilibrium constant for water dissociation and is inversely proportional to the hydroxide ion concentration. The two concentrations are related through the equation [H+][OH-] = 1.0 x 10^-14.

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Answer:

Explanation:

To find the concentration of hydroxide ions ([OH-]) in an aqueous solution, we can use the relationship between hydrogen ion concentration ([H+]) and hydroxide ion concentration in water at 25 °C, which is given by the equation:

[H+] x [OH-] = 1.0 x 10^-14

Given that the hydrogen ion concentration ([H+]) is 5.7 x 10^-10 (derived from the H* concentration provided), we can rearrange the equation to solve for [OH-]:

[OH-] = (1.0 x 10^-14) / [H+]

[OH-] = (1.0 x 10^-14) / (5.7 x 10^-10)

[OH-] ≈ 1.754 x 10^-5

Therefore, the concentration of hydroxide ions ([OH-]) in the given aqueous solution is approximately 1.754 x 10^-5.

Answer:

136.2295 grams of Ar

Explanation:

Simply multiply the moles by the grams. 3.41x39.95=136.2295.

Answer:

0.001464 M, or 1.464 × 10^(-3) M.

Explanation:

[H+] = 10^(-pH)

In this case, the pH of the acidic solution is 2.83. Plugging this value into the equation, we get:

[H+] = 10^(-2.83)

Using a calculator, we can find that 10^(-2.83) is approximately 0.001464.

Therefore, the concentration of hydrogen ions in the acidic solution is approximately 0.001464 M, or 1.464 × 10^(-3) M.

Explanation: heat is transferred from the hot water to the cold water until they reach the same temperature