There would be 10 moles of cobalt in 6.022 x [tex]10^{24[/tex] atoms of cobalt.
Avogadro's numberIn order to calculate the number of moles in a given number of particles, we need to divide the number of particles by Avogadro's number.
6.022 x [tex]10^{24[/tex] atoms of cobalt:
Number of moles = (number of particles) / Avogadro's numberNumber of moles = 6.022 x 10^24 / 6.022 x 10^23Number of moles = 10Therefore, there are 10 moles of cobalt in 6.022 x [tex]10^{24[/tex] atoms of cobalt.
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Boron trifluride reacts with hydrogen according to the following balaced equation 2BF3 3H2 = 2B 6HF suppose 40 grams of BF3 are mixed with 5 grams of H2 calculate the mass of HF in grams
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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It is an assignment question, so please check it properly to answer it and do use graph to explain it better!
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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42. The solubility of carbon dioxide in water is 0.161 g CO₂ in 100 mL of water at 20oC and 1.00 atm. A soft drink is carbonated with carbon dioxide gas at 5.50 atm pressure. What is the solubility of carbon dioxide in water at this pressure?
The solubility of carbon dioxide in water at the given pressure would be 8.45 g CO2 in 100 mL of water.
Solubility calculationAccording to Henry's law:
C = kH x P
where:
C is the concentration of the dissolved gas in the liquidP is the partial pressure of the gas above the liquidkH is the Henry's law constant for the specific gas-liquid systemTo 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:
0.161 g CO2 / 44.01 g/mol = 0.00366 mol CO20.00366 mol CO2 / 0.1 L = 0.0366 mol/LNow we can use Henry's law to calculate the solubility of carbon dioxide in water at 5.50 atm:
C = kH x PC = 0.0349 mol/L/atm x 5.50 atmC = 0.192 mol/LConcentration 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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Due today!
The diagram shows the potential energy changes for a reaction pathway.
A curved lined graph is shown. The y axis of the graph has the title Potential Energy. The x axis of the graph has the title Reaction Pathway. The graph curve begins at a lower level and ends at a slightly higher level. A vertical line labeled A, starting from the x axis till the beginning of the graph line is shown. A vertical line labeled B, starting from where the line labeled A ended, till the uppermost hump in the graph curve is shown. Another vertical line labeled C is shown from starting from the x axis till the point where the graph curve ends.
Part 1: Describe how you can determine the total change in enthalpy and activation energy from the diagram, and if each is positive or negative.
Part 2: Describe how the curve will look if the reaction was exothermic. Be sure to mention changes in the potential energies of the reactants and products and the sign changes of the enthalpy.
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)--
Why is eating prohibited in a science laboratory?
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:
Suggest reasons for and against the inclusion of hydrogen in the main groups of the periodic table
There are different reasons for and against the inclusion of hydrogen in the main groups of the periodic table.
Why hydrogen is included in the main groups of the periodic tableHydrogen 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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The thermite reaction
Fe2O3+2Al(s) → 2 Fe(s) + Al2O3(s)
used in some welding applications has the
following enthalpy and entropy changes at
298.15 K:
AH = -807.1 kJ/mol
AS = -0.0373 kJ/mol - K
Assuming AS and AH are constant, calculate
AG at 447 K.
Answer in units of kJ/mol.
The ΔG for the thermite reaction at 447 K is -816.894 kJ/mol.
How to calculate Gibbs free energy?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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Pls solve these 2 questions I will mark you as the brainliest
Which of the following is a way to conserve resources?
leave lights on
go one more trips
plan longer trips
take shorter showers
Which of the following ionic compounds would be expected to be insoluble in water?
NaOH
CaS
Ca(NO 3) 2
Na 2SO 4
KI
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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Calculate the mass of phosphoric(V) acid, H3PO4 , formed in the reaction
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
A driver needs to make a delivery to an office that is 30 km away. The driver has traveled for 45 minutes west down a straight road at 50 km/h. a. Has the driver traveled far enough to reach the office? Support your response. Input Field 1 of 2 Skip to input field
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