4. Calculate the number of moles in each of the following quantities:
a. 6.022 X 1024 atoms of cobalt

The solubility of carbon dioxide in water at the given pressure would be 8.45 g CO2 in 100 mL of water.  

According to Henry's law:

C = kH x P

where:

To solve this problem, we need to use the Henry's law constant for carbon dioxide in water at 20 [tex]^oC[/tex], which is 0.0349 mol/L/atm. Let's convert the given solubility from mass/volume units to molar concentration units as follows:

Now we can use Henry's law to calculate the solubility of carbon dioxide in water at 5.50 atm:

Concentration from mol/L to mass/volume units:

0.192 mol/L x 44.01 g/mol = 8.45 g CO2 / 100 mL

Therefore, the solubility of carbon dioxide in water at 5.50 atm pressure is approximately 8.45 g CO2 in 100 mL of water at 20 [tex]^oC[/tex].

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The mass of the phosphoric(V) acid, H₃PO₄ , is formed in the reaction is 7 g.

The chemical equation is as :

PCl₅ + H₂O  --->   H₃PO₄ + HCl

The mass of the PCl₅ = 15 g

The moles of the PCl₅ = mass / molar mass

The moles of the PCl₅ = 15 / 208.24

The moles of the PCl₅ = 0.0720 mol

The 1 mol of the PCl₅ forms the 1 mole of the H₃PO₄

The moles of the H₃PO₄ = 0.0720 mol

The mass of the H₃PO₄ = moles ×  molar mass

The mass of the H₃PO₄ = 0.0720 × 97.99

The mass of the H₃PO₄ = 7 g.

The mass of the Phosphoric acid, H₃PO₄ is 7 g.

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This question is incomplete, the complete question is :

Calculate the mass of phosphoric(V) acid, H3PO4 , formed in the reaction. The mass of the PCl₅ is 15 g.

PCl₅ + H₂O  --->   H₃PO₄ + HCl

None of these compounds are insoluble in water which means that all of the compounds are soluble in water.

Solubility rules are a set of conditions that can be used to easily determine the possible results of a mixture between a solute and a solvent.

Generally, it defines the terms by which the solution will be saturated, or whether precipitation will occur. In addition to these conditions, solubility also depends on the pressure and temperature where the mixture is formed.

Thus all of these compounds are soluble in water.

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Answer: Yes

Explanation:

Convert [tex]30km[/tex] to [tex]min.[/tex] by using the conversion factor that is given which is [tex]\frac{50km}{1hr}[/tex].

[tex]30km/\frac{1hr}{50km}/\frac{60min}{1hr}=36min[/tex]

This means it would take 36 minutes for the driver to reach 30 km

The ΔG for the thermite reaction at 447 K is -816.894 kJ/mol.

The Gibbs free energy change (ΔG) of a reaction can be calculated using the equation:

ΔG = ΔH - TΔS

where ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin.

First, calculate ΔG at 298.15 K:

ΔG(298.15 K) = -807.1 kJ/mol - 298.15 K x (-0.0373 kJ/mol-K)

ΔG(298.15 K) = -807.1 kJ/mol + 11.146 kJ/mol

ΔG(298.15 K) = -795.954 kJ/mol

Using the following equation to calculate ΔG at 447 K:

ΔG(447 K) = ΔG(298.15 K) + ΔH(T2-T1)/T2 + ΔS ln(T2/T1)

where T1 = 298.15 K and T2 = 447 K

ΔG(447 K) = -795.954 kJ/mol + (-807.1 kJ/mol)(447 K - 298.15 K)/447 K + (-0.0373 kJ/mol-K)ln(447 K/298.15 K)

ΔG(447 K) = -795.954 kJ/mol - 24.902 kJ/mol + 3.962 kJ/mol

ΔG(447 K) = -816.894 kJ/mol

Therefore, the ΔG for the thermite reaction at 447 K is -816.894 kJ/mol.

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Answer:

Because the food can fall and make a mess not only that but it can go into a experiment or something the scienctist are working on and ruin it, cause a reaction.

Explanation:

There are different reasons for and against the inclusion of hydrogen in the main groups of the periodic table.

Hydrogen has properties that justify its inclusion in the main groups of the periodic table, including its chemical behavior, abundance in the universe, and the structure of its atoms.

However, its ambiguity in properties, existence in three isotopes, and lower abundance on Earth can be arguments against its classification. The unique chemistry and abundance of hydrogen in the universe support its inclusion in the periodic table.

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Part 1. The combustion of methane ([tex]CH_4[/tex]) to form carbon dioxide ([tex]CO_2[/tex]) and water ([tex]H_2O[/tex]) is an exothermic reaction. Part 2. It means that it releases energy.

1. The potential energy change for a reaction pathway in this reaction can be calculated by comparing the potential energy of the reactants ([tex]CH_4[/tex] and [tex]O_2[/tex]) to the potential energy of the products ([tex]CO_2[/tex] and [tex]H_2O[/tex]). In this case, the potential energy of the reactants is higher than the potential energy of the products. 2. It means that energy is released during the reaction. The potential energy change for a reaction pathway in the combustion of methane is approximately -890 kJ/mol. This energy release is what makes combustion reactions useful for heating and energy production.

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--The complete Question is, What is the potential energy change for a reaction pathway in the combustion of methane (CH4) to form carbon dioxide (CO2) and water (H2O)--

The instantaneous rate of reaction at 17 minutes is approximately -0.178 mol dm⁻³

To find the instantaneous rate of reaction at 17 minutes, we can use the concept of differential calculus and estimate the slope of the tangent line at t=17 on the graph of rate versus time.

To do this, we can use the formula for the slope of a line

slope = (change in y) / (change in x)

In this case, the "y" values are the rates of reaction and the "x" values are the times. We want to find the slope at t=17, so we can choose two points that are very close to t=17, such as t=15 and t=20. Then, we can use these values to estimate the slope at t=17

slope = (rate at 20 min - rate at 15 min) / (20 min - 15 min)

slope = (0.135 - 0.223) / (20 - 15)

slope = -0.178

This slope represents the instantaneous rate of reaction at t=17. However, since it has a negative value, it means that the rate of reaction is decreasing at t=17.

Therefore, the instantaneous rate of reaction at 17 minutes is approximately -0.178 mol dm⁻³

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The mass of HF produced from the given amounts of BF₃ and H₂ is 99.5 grams.

To solve this problem, we need to use stoichiometry to determine the amount of HF produced from the given amounts of BF₃ and H₂.

From the balanced equation;

2 BF₃ + 3 H₂ → 2 B + 6 HF

We can see that 2 moles of BF₃ react with 3 moles of H₂ to produce 6 moles of HF. This means that the ratio of BF₃ to HF is 2:6, or 1:3.

First, we need to calculate the number of moles of BF₃ and H₂ from their given masses and molar masses;

n(BF₃) = m(BF₃) / M(BF₃) = 40 g / 67.81 g/mol = 0.59 mol

n(H₂) = m(H₂) / M(H₂) = 5 g / 2.02 g/mol = 2.48 mol

Next, we can determine which reactant is limiting by comparing their mole ratios. The ratio of BF₃ to H₂ is 1:1.67, which means there is not enough H₂ to react with all of the BF₃. This makes H₂ the limiting reactant.

To find the number of moles of HF produced, we use the mole ratio from the balanced equation;

n(HF) = n(H₂) × (6 mol HF / 3 mol H₂) = 2.48 mol × 2

= 4.96 mol

Finally, we calculate the mass of HF produced from its number of moles and molar mass;

m(HF) = n(HF) × M(HF) = 4.96 mol × 20.01 g/mol

= 99.5 g

Therefore, the mass of HF is 99.5 grams.

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