What is the amount of pure substance in 10. 7 g of aluminum? Assume the molar mass of aluminum is 26. 98 g/mol

Answer:

the formula for sulfide ion is S²⁻, and the formula for calcium ion is Ca²⁺. When combined, they form calcium sulfide with the chemical formula CaS.

Explanation:

Answer:

6 moles of H2O

Explanation:

We need the balanced equation for this reaction.

   C2H2 + O2 = CO2 + H2O

   2C2H2 + 5O2 = 2CO2 + 4H2O   [Balanced reaction]

The balanced reaction tells us that we should expect to obtain 4 moles of water, H2), for every 2 moles of C2H2 (ethylene).

That is a ratio:  (4 moles H2O)/(2 moles C2H2) or (2/1)(moles H2O/moles C2H4)

If we start with 3 moles of C2H4:

 (3 moles C2H4)*((2 moles H2O)/(1 moles C2H2)) = (2*3/1) moles H2O [the moles C2H2 cancel]

(2*3/1) moles H2O = 6 moles of H2O

This means that for every liter of water, there is 6.15 moles of the substance dissolved in it.

Substance is a concept that refers to a physical material or thing that has mass and occupies space. It is a fundamental concept of physics that applies to all physical and visible things in the universe. In philosophy, substance is a primary category of ontology that refers to the physical or material existence of things.

The molarity of the saturated solution can be calculated by dividing the solubility (360 g/L) by the molar mass of the substance (58.5 g/mol).
The molarity of the saturated solution is thus 6.15 mol/L.
This means that for every liter of water, there is 6.15 moles of the substance dissolved in it.

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Answer:

We can begin by using the balanced chemical equation for the reaction between carbon monoxide and oxygen:

2CO + O2 -> 2CO2

This tells us that 2 volumes of CO react with 1 volume of O2 to produce 2 volumes of CO2. Since we have equal volumes of CO and O2, the limiting reactant will be the one that requires more volume, which is the CO:

2 volumes of CO + 1 volume of O2 -> 2 volumes of CO2

Using the ideal gas law, we can convert the volumes of CO and O2 at STP (standard temperature and pressure, which is 0°C and 1 atm) to moles:

n(CO) = V(CO) / Vm = 60 cm3 / 22.4 L/mol = 0.00268 mol

n(O2) = V(O2) / Vm = 30 cm3 / 22.4 L/mol = 0.00134 mol

Since 2 volumes of CO react with 1 volume of O2, we can say that the reaction will use up 2 x 0.00134 = 0.00268 mol of O2. Since we have 0.00134 mol of O2 initially, this means that all of the O2 will be used up in the reaction, leaving none in the residual gases. The reaction will also produce 2 x 0.00268 = 0.00536 mol of CO2.

Using the ideal gas law again, we can convert the volume of CO2 produced to a volume at STP:

V(CO2) = n(CO2) x Vm = 0.00536 mol x 22.4 L/mol = 0.120 cm3

Therefore, the volume of residual gases after the reaction is:

V(residual) = V(total) - V(CO) - V(O2) - V(CO2) = 90 cm3 - 60 cm3 - 30 cm3 - 0.120 cm3 = 0.880 cm3

The percentage by mass of water in the hydrate [tex]CuSO^{4}.5H^{2}O[/tex] is 47.6%.

The percentage composition of each of the following compounds are

1.a. NaClO:

The molar mass of NaClO =

22.99 g/mol + 35.45 g/mol + 15.99 g/mol = 74.43 g/mol

Percentage composition of Na =

(22.99 g/mol / 74.43 g/mol) *  100 \ percent \ = 30.9percent

Percentage composition of Cl =

(35.45 g/mol / 74.43 g/mol) *  100percent \ =  47.6percent

Percentage composition of O =

(15.99 g/mol / 74.43 g/mol) * 100percent \ =  21.5percent

Therefore, the percentage composition of NaClO is approximately:

30.9% Na, 47.6% Cl, and 21.5% O.

1.b. [tex]Al^{2}(SO^{3})^{3}[/tex]:

The molar mass of [tex]Al^{2}(SO^{3})^{3}[/tex] = [tex]2 * (26.98 g/mol) + 3 * (32.06 g/mol + 3 * 16.00 g/mol) = 342.14 g/mol[/tex]

Percentage composition of Al =

[tex](2 * 26.98 g/mol / 342.14 g/mol) * 100 \ = 3.99[/tex]%

Percentage composition of S =

[tex](3 * 32.06 g/mol / 342.14 g/mol) * 100 \ = 8.99[/tex]%

Percentage composition of O =

[tex](9 * 16.00 g/mol / 342.14 g/mol) * 100 \ = 10.47[/tex]%

Therefore, the percentage composition of [tex]Al^{2}(SO^{3})^{3}[/tex] is approximately:

3.99% Al, 8.99% S, and 10.47% O.

1.c. [tex]C^{2}H^{5}COOH[/tex]:

The molar mass of [tex]C^{2}H^{5}COOH[/tex] = [tex]2 * (12.01 g/mol) + 6 * (1.01 g/mol) + 12.01 g/mol + 16.00 g/mol = 60.05 g/mol[/tex]

Percentage composition of C =

[tex](2 * 12.01 g/mol / 60.05 g/mol) * 100 \ = 40.0[/tex]%

Percentage composition of H =

[tex](6 * 1.01 g/mol / 60.05 g/mol) * 100\ \ = 10.1[/tex]%

Percentage composition of O = [tex](2 * 16.00 g/mol + 12.01 g/mol / 60.05 g/mol) * 100 \ = 49.9[/tex]%

Therefore, the percentage composition of [tex]C^{2}H^{5}COOH[/tex] is approximately:

40.0% C, 10.1% H, and 49.9% O.

1.d. [tex]BeCl^2[/tex]:
The molar mass of [tex]BeCl^2[/tex] = [tex]9.01 g/mol + 2 * 35.45 g/mol = 79.91 g/mol[/tex]

Percentage composition of Be = [tex](9.01 g/mol / 79.91 g/mol) * 100 \ = 11.3[/tex]%

Percentage composition of Cl =

[tex](2 * 35.45 g/mol / 79.91 g/mol)* 100 \ = 88.7[/tex]%

Therefore, the percentage composition of [tex]BeCl^2[/tex] is approximately:

11.3% Be and 88.7% Cl.

2. [tex]CuSO^{4}.5H^{2}O[/tex]:

Mass of CuSO4.5H2O = 8.40 g

Mass of anhydrous [tex]CuSO^{4}[/tex] = mass of [tex]CuSO^{4}.5H^{2}O[/tex] - mass of water

Mass of water = mass of [tex]CuSO^{4}.5H^{2}O[/tex] - mass of anhydrous CuSO4

[tex]= 8.40 g - 4.40 g= 4.00 g[/tex]

So, the mass of water in the hydrate is 4.00 g.

To determine the percentage by mass of water in the hydrate, we need to divide the mass of water by the mass of the entire compound ([tex]CuSO^{4}.5H^{2}O[/tex]) and multiply by 100%:

Percentage by mass of water = (mass of water/mass of [tex]CuSO^{4}.5H^{2}O[/tex]) x 100%

[tex]= (4.00 g/8.40 g) * 100= 47.6[/tex]%

Therefore, the percentage by mass of water in the hydrate [tex]CuSO^{4}.5H^{2}O[/tex] is 47.6%.

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If 9.7 mol of CO₂ are produced, 14.55 moles of O₂ were reacted.

In chemistry, a mole is a unit of measurement that represents an amount of substance. This number of entities is known as Avogadro's number, which is approximately equal to 6.02 x 10^23.

The mole is a convenient unit for measuring the amount of substance in chemical reactions, because it allows chemists to predict and calculate the quantities of reactants and products involved in a chemical reaction. For example, if we know the balanced chemical equation for a reaction, we can use the stoichiometry of the reaction to determine the amount of reactants needed to produce a given amount of product, or the amount of product that can be produced from a given amount of reactants.

The balanced chemical equation for the combustion of ethylene (C2H4) in oxygen is:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

From the balanced equation, we can see that 3 moles of O₂ react with 1 mole of C₂H₄ to produce 2 moles of CO₂. Therefore, the number of moles of O₂ that reacted can be calculated as:

moles of O₂ = (moles of CO₂ produced) × (3/2)

moles of O₂ = 9.7 mol × (3/2)

moles of O₂ = 14.55 mol

Therefore, 14.55 moles of O₂ were reacted.

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The pH of the 0.30 M solution of sodium formate is 2.13.
The correct answer is E. 2.13.

Formic acid dissociates in water to form hydrogen formate (HCOO-) and a proton (H+).

Since the Ka of HCOOH is [tex]1.8 * 10^{-4}[/tex], the Kb of HCOO- is equal to Ka x Kb = ([tex]1.8 * 10^{-4}[/tex])([tex]1.0 * 10^{-14}[/tex]) = [tex]1.8 * 10^{-18}[/tex].

The ionization of sodium formate (NaHCOO) can be represented as:

[tex]NaHCOO + H^2O < = > HCOO^- + Na^+ +[/tex] [tex]H^3O^+[/tex]

We can use the Kb of HCOO- to calculate the equilibrium concentrations of the species in the solution.

[HCOO-] = [NaHCOO] = 0.30 M

[Na+] = [[tex]H^3O^+[/tex]] = 0

Kb = [HCOO-][[tex]H^3O^+[/tex]]/[NaHCOO]

1.8 x 10-18 = (0.30)(x)/(0.30)

x = 1.8 x 10-18

The pH of the solution can then be calculated using the equation:

pH = -log[[tex]H^3O^+[/tex]]

pH = -log(1.8 x 10-18)

pH = 2.13

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Question should be like

Calculate the pH of a 0. 30 M solution of sodium formate (NaHCOO) given that the Ka of formic acid (HCOOH) is 1. 8 x 10^-4.

A. 8. 61

B. 10. 26

C. 11. 87

D. 5. 39

E. 2. 13

T = 23 C + 273 = 296 K V(H2) = (1.7 mol*62360 mmHg/mol*K*296 /850 mmHg) = 36917.1 ml = 36.9 L if volume, per liters, of H2 gas is created at 26 C and 800 mmHg.

For instance, hydrogen is not harmful. Also, as hydrogen is that much lighter than the atmosphere, it evaporates quickly when it is discharged, enabling the fuel to disperse relatively quickly in the event of a leak. To ensure its safe use, some of hydrogen's features necessitate additional technical restrictions.

When two atoms of hydrogen join forces to form a hydrogen molecule, the result is a gas known as H2. Molecular hydrogen is another name for H2. There are two protons & two electrons in it. Hence, it has a neutral charge and is stable, making it the most prevalent type of hydrogen.

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The balanced equation for the reaction would be 2AlPO4 + 3CuCrO4 → Al2(CrO4)3 + 3Cu3(PO4)2.

The formulas of the reactants are:

The formulas of the products are:

The balanced equation indicates that 2 moles of Aluminum Phosphite react with 3 moles of Copper (II) Chromate to produce 1 mole of Aluminum Chromate and 3 moles of Copper (II) Phosphate.

Thus: 2AlPO4 + 3CuCrO4 → Al2(CrO4)3 + 3Cu3(PO4)2

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Answer:

From the balanced net ionic equation, we can see that 1 mole of AgNO3 reacts with 1 mole of CaCl2 to produce 1 mole of AgCl. Therefore, we can use the stoichiometry of the reaction to calculate the amount of AgNO3 needed to react completely with the given amount of CaCl2.

First, we need to calculate the number of moles of CaCl2 in 76.5 mL of 0.391 M solution:

moles of CaCl2 = concentration x volume

moles of CaCl2 = 0.391 M x 0.0765 L

moles of CaCl2 = 0.0299 mol

Since 1 mole of AgNO3 reacts with 1 mole of CaCl2, we need 0.0299 mol of AgNO3 to react completely with the CaCl2. To calculate the volume of 0.164 M AgNO3 solution containing this amount of AgNO3, we can use the formula:

moles = concentration x volume

Rearranging the formula, we get:

volume = moles / concentration

Substituting the values, we get:

volume = 0.0299 mol / 0.164 M

volume = 0.1823 L

Converting to milliliters:

V(AgNO3) = 182.3 mL

Therefore, 182.3 mL of 0.164 M AgNO3 solution are needed to react completely with 76.5 mL of 0.391 M CaCl2 solution.

The various types of chemical reaction are Combination Reaction,Decomposition Reaction,Displacement Reaction,Double Displacement Reaction, and Precipitation Reaction.

A chemical reaction is defined as the reaction that involves the combination of two or more substances leading to the formation of a new substance called the product of the reaction.

The various types of chemical reaction include the following:

There are three parts of a chemical reaction which is the reactant part, the arrow and the product part.

That is;

A. + B ---------> D + E.

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The energy required to increase 1.000 kg of water ice between 0 °C °C through 79.8 °C °C is equivalent to the energy required to melt one pound of ice (334 kJ).

No surface temp change happens from heat exchange if ice caps melt and has become water ice (i.e., that during phase transition). Imagine about water melting from stalactites that were already dripping on a sun-warmed roof. Alternatively, water boils in an ice bucket cooled by relatively low surroundings.

To melt, the iceberg will take in all the heat that is available. The water stays at zero degrees Fahrenheit until all of the ice has dissolved as when the ice has melted, it changes becomes saltwater at that temperature.

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All the statements are about the acidity of compounds, of which statement 1, 3 & 4 are true and statement 2 is false.

Statement 1: "the strength of an acid depends primarily on the conjugate base's ability to support a negative charge" is true because, a strong acid has a weak conjugate base, which means it is less able to support a negative charge.

Statement 2: "induction is long range electronegativity, and the presence of electronegative atoms improves acidity of distant H" is false.

Because, induction is the ability of an electronegative atom or group to withdraw electron density from a neighbouring atom or group, leading to an increase in acidity.

However, this effect is not long range, and typically only affects neighbouring atoms.

Statement 3: "H bonded to larger atoms are more acidic due to poor orbital overlap, and polarizability of large orbitals" is true.

Because larger atoms have larger orbitals, which leads to poor orbital overlap and weaker bonds with hydrogen.

This makes it easier for the hydrogen to dissociate and increases the acidity of the compound.

Statement 4: "increasing the % of s-orbital character in a bond increases acidity" is true.

S-orbitals are closer to the nucleus and have a lower energy than p-orbitals, leading to stronger bonds and increased acidity.

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The synthesis of carboxylic acid can be done by hydrolysing the nitrile which is substituted in place of halide by using cyanide ion. The structure that contains a nitrile is OCH2C≡N.

A nitrile is an organic compound that contains a carbon triple bonded to a nitrogen atom.

The general formula for a nitrile is R–C≡N, where R is an alkyl or aryl group. In the structure OCH2C≡N, the OCH2 group is the alkyl group and the C≡N is the nitrile functional group. Therefore, this structure contains a nitrile.

To synthesize a carboxylic acid from an alkyl halide, you can first convert the alkyl halide to a nitrile through a substitution reaction with a cyanide ion. The nitrile can then be hydrolyzed to form a carboxylic acid. The overall reaction is:

R–X + CN– → R–C≡N → R–COOH

Where R is an alkyl or aryl group, X is a halogen, and CN– is a cyanide ion.

In conclusion, the structure that contains a nitrile is OCH2C≡N and it can be used to synthesize a carboxylic acid from an alkyl halide through a substitution reaction followed by hydrolysis.

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The new pressure of the gas with a solubility of 24 g/L will be222 kPa.

The solubility of the gas in the liquid is directly proportional to the partial pressure of a gas above the liquid. This relationship is described by Henry's Law, which states that the solubility of a gas is proportional to its partial pressure:

C = kP

where C is the concentration of the gas in the liquid (in units of g/L), P is the partial pressure of the gas above the liquid (in units of kPa), and k is the proportionality constant.

If we know the solubility of the gas at one pressure, we can use Henry's Law to calculate the solubility at a different pressure. For example, if the solubility of the gas is 48 g/L at 444 kPa, and we want to know the solubility at a new pressure, we can use the following formula:

P₁ / P₂ = C₁ / C₂

where P₁ is the initial pressure, P₂ is the new pressure, C₁ is the initial concentration (solubility) at P₁, and C₂ is the new concentration (solubility) at P₂.

Rearranging this formula, we get:

C₂ = C₁ x (P₂ / P₁)

Substituting the given values, we have:

C₁ = 48 g/L (solubility at 444 kPa)

P₁ = 444 kPa

C₂ = 24 g/L (desired solubility at new pressure)

P₂ = ?

Solving for P₂, we get:

P₂ = (C₂ x P₁) / C₁

P₂ = (24 g/L x 444 kPa) / 48 g/L

P₂ = 222 kPa

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A. The limiting reactant is NH₃

B. The mass of NO formed is 44.1 g

C. The mass of H₂O formed is 39.7 g

First, we shall determine the mass in 4 moles of O₂. Details below:

Mole = mass / molar mass

4 = Mass of O₂ / 32

Cross multiply

Mass of O₂ = 4 × 32

Mass of O₂ = 138 g

Finally, we shall determine the limiting reactant. Details below:

4NH₃ + 5O₂ -> 4NO + 6H₂O

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂

Therefore,

25 g of NH₃ will react with = (25 × 160) / 68 = 58.8 g of O₂

From the above calculation, we can see that only 58.8 g of O₂ out of 138 g is needed to react with 25 g NH₃.

Thus, the limiting reactant is NH₃

The mass of NO formed can be obtained as illustrated below:

4NH₃ + 5O₂ -> 4NO + 6H₂O

From the balanced equation above,

68 g of NH₃ reacted to produce 120 g of NO

Therefore,

25 g of NH₃ will react to produce = (25 × 120) / 68 = 44.1 g of NO

Thus, the mass of NO formed is 44.1 g

The mass of H₂O formed can be obtained as illustrated below:

4NH₃ + 5O₂ -> 4NO + 6H₂O

From the balanced equation above,

68 g of NH₃ reacted to produce 108 g of H₂O

Therefore,

25 g of NH₃ will react to produce = (25 × 108) / 68 = 39.7 g of H₂O

Thus, the mass of H₂O formed is 39.7 g

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Answer: In the chlor alkali process, sodium hydroxide is prepared from common salt (sodium chloride). The byproducts are hydrogen gas and chlorine gas. Electrolysis of aqueous sodium chloride (brine solution) in Castner-kellner cell gives sodium hydroxide, chlorine gas and hydrogen gas.

Answer: Useful substances are obtained from common salt by the process of electrolysis

Answer:

Hence, C121H199O100 represents the compound's empirical formula.

Explanation:

We must ascertain the ratio of the number of atoms of each element in the compound in order to derive the empirical formula. To accomplish this, we can divide the mole ratio by the least number of moles after converting the masses of each element to moles.

The elements' molar masses are as follows:

12.01 g/mol for carbon

1.01 g/mol for hydrogen

16.00 g/mol for oxygen

When we convert the masses to moles, we obtain:

48.38 g / 12.01 g/mol = 4.03 mol of carbon

6.74 g / 1.01 g/mol of hydrogen equals 6.67 mol

53.5 g / 16.00 g/mol = 3.34 mol for oxygen

3.34 mol of oxygen is the least amount of moles. If you divide the total moles of each element by 3.34 mol, you get:

Carbon: 4.03 mol/3.3 mol = 1.21 mol Hydrogen:6.67 moles/3.34 moles equals 1.99

3.34 mol / 3.34 mol = 1.00 for oxygen.

By multiplying each ratio by 100 to obtain whole integers, we may simplify the ratios, which are around 1.21:1.99:1.00. This results in a ratio of roughly 121:199:100.

We divide each number in the ratio by their greatest common factor (GCF) to arrive at the empirical formula. 121, 199, and 100 have a GCF of 1. By dividing by 1, we get:

Carbohydrate: 121 / 1 = 121

199 / 1 = 199 for hydrogen.

100 / 1 Equals 100 for oxygen.

Hence, C121H199O100 represents the compound's empirical formula.

Answer:

First, I will convert from grams to moles. This will give me approximately 4.032 moles of carbon, 6.740 moles of hydrogen, and 3.344 moles of oxygen. Then, I will calculate the mole ratio, using whole numbers. I must also identify the least amount of moles in an element for this step, which is oxygen at 3.344 moles.

Oxygen = 3.344/3.344 = 1

Carbon = 4.032/3.344 = 1

Hydrogen = 6.740/3.344 = 2

The empirical formula will be COH2.

Now that we have the molar mass of the molecular formula, we can calculate for the molecular formula. First, we will begin by discovering the molar mass of the empirical formula, which will be C molar mass x 1 + O molar mass x 1 + H molar mass x 2. That will be equal to 60.949. After that, we can divide the molecular mass by the empirical formula molar mass, giving us approximately 3. Now we will multiply the empirical formula by 3, giving us our molecular formula, which will be equal to C3O3H6.

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North America is experiencing spring during the position when the Northern Hemisphere is tilted towards the Sun, which occurs around March 20th to 21st, known as the Vernal Equinox.

The point in the Earth's orbit around the Sun when the tilt of axis is neither towards nor away from the Sun is known as the Vernal Equinox, also known as the Spring Equinox. Because of this arrangement, the length of day and night is almost identical over the whole globe. This occasion, which occurs in North America around March 20–21, heralds the start of the spring season. The Northern Hemisphere begins to tilt towards the Sun around this time, lengthening the day and raising temperatures. Agriculture, migratory patterns, and different cultural holidays are significantly impacted by this change in the Earth's position.

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Answer:

10. methanol would be more.

11. methanol would be more.

12. methanol would remain the same.

13. CO would be less.

Explanation:

10. Methanol increases because the forward reaction is favoured with increase in temperature and thus it is endothermic.

11. Methanol increases because the forward reaction is favoured with decrease in volume thus increasing pressure decreases the volume of the vessel.

12.Methanol remains the same because the hydrogen added is its raw material but the CO is acting as a limiting reactant for the formation of methanol.

13. Carbon monoxide becomes less because it will rapidly combine with hydrogen added to form methanol and then later be a limiting reactant when hydrogen is still present

Answer: sp3

Explanation:

The correct answer is that Friedel-Crafts reactions can not be done on moderately or strongly deactivated ring systems.

This is a limitation of Friedel-Crafts reactions because they require an activated ring system in order to proceed.

Deactivated ring systems do not have the necessary electron density to facilitate the reaction, and therefore the reaction will not occur.

Friedel-Crafts alkylation and acylation reactions are both types of electrophilic aromatic substitution reactions that involve the formation of a carbon-carbon bond between an aromatic ring and an alkyl or acyl group.

These reactions are commonly used in organic synthesis to introduce new functional groups onto an aromatic ring.

However, they do have other limitations including the fact that Friedel-Crafts alkylation substrates can undergo rearrangement,

Friedel-Crafts acylation often leads to polyacylated products,

and Friedel-Crafts alkylation often leads to polyalkylated products.

These limitations must be taken into consideration when designing a synthetic route that involves a Friedel-Crafts reaction.

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the half-life of the reaction is 1.29 s (to three significant figures).

The half-life (t1/2) of a first-order reaction is given by the formula:

t1/2 = ln(2) / k

where k is the rate constant of the reaction.

In this case, the rate law for the reaction is:

rate = k[A]3[B]2[C]

where [A], [B], and [C] are the concentrations of A, B, and C, respectively.

Since the reaction is first order with respect to A, the concentration of A at any time t is given by:

[A]t = [A]0 x

[tex] {e}^{ - kt} [/tex]

where [A]0 is the initial concentration of A.

Given that the initial concentration of A is 0.867 mol/L and the rate constant is 0.538 , we can use the formula for the half-life to calculate the time required for the concentration of A to decrease by half:

t1/2 = ln(2) / k = ln(2) / 0.538 = 1.29 s

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Answer:

c

Explanation:

energy neither created nor distroyed

Answer:

The answer is C

Explanation:

We learned so many times about law conservation energy,states that"Energy neither created or destroyed but can change its form to other."

E.g If we rub our palm,it changes from kinetic energy to heat energy.

Answer: First, we need to calculate the total fuel consumption per day for all 380 SSTs:

Fuel consumption per hour: 5,320 gallons

Number of planes: 380

Hours per day: 14

Total fuel consumption per day = 5,320 x 380 x 14 = 28,190,400 gallons

Next, we need to convert gallons to metric tons of jet fuel:

1 gallon = 0.00378541 metric tons

28,190,400 gallons = 106,698.89 metric tons

Now, we can calculate the total crude oil needed to produce this amount of jet fuel:

1 ton of jet fuel = 7,000 kilograms of crude oil

106,698.89 tons of jet fuel = 746,892,230 kilograms of crude oil

To convert kilograms to metric tons:

1 metric ton = 1,000 kilograms

746,892,230 kilograms = 746,892.23 metric tons

Finally, we can calculate the percentage of crude oil production used for SST fuel:

World crude oil production in 1990 = 4.02 x 10^9 metric tons

Percentage of crude oil used for SST fuel = (746,892.23 / 4.02 x 10^9) x 100

= 0.0186%

Therefore, approximately 0.0186% of the crude oil production in 1990 was used for fuel for the SSTs.

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A student measures the molar solubility of nickel(II) cyanide in a water solution to be 2. 00×10-8 M. The Ksp of nickel(II) cyanide is [tex]8 \times10^{-24}[/tex]

The solubility product constant, Ksp, serves as the equilibrium constant for solids that dissolve in water. The level of solute dissolution in solution is what it stands for. The more soluble a material is, the higher its Ksp value.

The following equation may be used to get the solubility product constant (Ksp) for nickel(II) cyanide from the molar solubility:

[tex]Ni(CN)_2(s)[/tex] ⇌ [tex]Ni_2+(aq) + 2CN^-(aq)[/tex]

[tex]Ksp = [Ni_2^+][CN^-]^2[/tex]

As per the given information,  

the molar solubility of nickel(II) cyanide is given as [tex]2.00\times 10^{-8}[/tex]M, the concentrations of [tex]Ni_2^+[/tex] and [tex]CN^-[/tex] ions are also [tex]2.00 \times 10^{-8}[/tex] M.

We have to find the Ksp of nickel(II) cyanide.

Hence, the Ksp can be calculated as:

Ksp = [tex](2.00\times10^{-8})(2.00\times10^{-8})^{2}[/tex]

Ksp =[tex]8.00\times10^{-24}[/tex]

Hence, the Ksp of nickel(II) cyanide is [tex]8.00\times10^{-24}[/tex]

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Answer:

Answer :

The compounds are separated by using a suitable filtration technique.

Explanation:

Explanation:

The compounds are separated by using a suitable filtration technique. NaCl remains in the filtrate, but charcoal remains on the filter paper. Crystals of NaCl can be obtained by the method of evaporation.

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