The Rydberg formula can calculate the wavelengths of light emitted by hydrogen atoms. However, with some modifications, it can also be used to calculate the wavelengths of light emitted by atoms of other elements. Hence, the initial energy level is n = 3
Rydberg equation:
1/λ = f/c = R(1/m² - 1/n²)
where c is the speed of light, f is frequency, λ is the wavelength, R is Rydberg constant, and n and m are the quantum numbers of the energy levels.
The initial energy level has quantum number n.
The final energy level is a ground state with the quantum number m = 1.
2.924 x 1015/2.998 x 108 = 1.097 x 107 x (1/12 - 1/n²)
(1/12 - 1/n²) = 0.8891
1/n² = 0.1109
=> n²
= 9
=> n = 3
Thus, the initial energy level is n = 3
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The electron in the hydrogen atom was in the 9th energy level (n = 9) before emitting the photon.
Explanation:When a hydrogen atom transitions from an excited state to the ground state, it emits a photon with a certain frequency. In this case, the frequency is given as ν = 3.084 x 10^15 s^-1. We can use the equation ν = E/h, where E is the energy of the photon and h is Planck's constant, to find the energy of the emitted photon. Once we have the energy, we can determine the energy level the electron was in before the photon was emitted.
Using the energy-frequency relationship and Planck's constant, we have:
E = hν = (6.626 x 10^-34 J s)(3.084 x 10^15 s^-1) ≈ 2.041 x 10^-18 J
From this energy, we need to find the corresponding energy level. The energy levels of hydrogen are given by the formula: E = -13.6 eV/n^2, where E is the energy of the level and n is the principal quantum number. Rearranging the formula, we have:
n^2 = -13.6 eV/E ≈ -13.6 eV/(2.041 x 10^-18 J) ≈ -6.662 x 10^17
Taking the square root of both sides, we find:
n ≈ -8.159 x 10^8
Since n must be a positive integer, we can conclude that the electron was in the 9th level (n = 9) before emitting the photon.
1. a substance that changes the rate of the reaction, and is not consumed in the reaction solution
2. a uniform mixture of molecules or ions of one substance in another solvent
3. the substance dissolved in a solution solute
4. that component of a solution which dissolves the solute catalyst
Answer:
1. Catalyst
2. Solution
3. Solute
4. Solvent
Explanation:
1. A catalyst increases reaction rate but is not used up or consumed in the reaction
Example of a catalyst found in nature are Enzymes
2. Solution, an homogeneous mixture of two or more substances solutes and solvents
3. Solute, the component of a solution dissolved in the solvent
4. Solvent, the component of a solution that dissolves the solute
arrange the following solutions in order of decreasing freezing point: 0.10m Na3PO4; 0.35m NaCl; 0.20m MgCl2; 0.15m C6H12O6; 0.15m CH3COOH.
Answer:
0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 > 0.20 m MgCl2 > 0.35 m NaCl
Explanation:
Step 1:
0.10m Na3PO4
Na3PO4 → 3Na+ + PO4^-3-
⇒ van't Hoff factor is 3 + 1 = 4
0.10 m * 4 = 0.40
0.35m NaCl
NaCl → Na+ + Cl-
⇒ van't Hoff factor = 1+1 = 2
0.35 * 2 = 0.70
0.20m MgCl2
MgCl2 → Mg^2+ + 2Cl-
⇒ Van't Hoff factor = 1+2 = 3
0.20 * 3 = 0.60
0.15m C6H12O6
⇒ for non-ionic compounds in solution, like glucose (C6H12O6) , the van't Hoff factor is 1. They do not dissociate in water.
0.15 * 1 = 0.15
0.15m CH3COOH.
CH3CO2H ⇄ CH3CO2− + H+
⇒ Van't hoff factor ≈ 1<x<2
0.15 * 2 = 0.30
Larger the concentration of the concentration of the particles, smaller the freezing point.
0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 > 0.20 m MgCl2 > 0.35 m NaCl
The order of decreasing freezing point is -
0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,
The Freezing Point DepressionIt is calculated as:
dT = i Kf m
m = The concentration values in molalities
Kf = the cryoscopic constant for water (-1.86, the same for all solutions)
i = the Van’t Hoff factor, which is the number of ion particles per individual molecule/formula of solute
Solution:
So you have to arrange in increasing order of the product i·m
0.15 m C6H12O6
im = 1 x 0.15 = 0.15 (undissociated, i=1)
0.15 m CH3COOH
im =1.1 x 0.15 = 0.17 (partially dissociated) ( 1<i<2)
0.10 m Na3PO4
im = 4 x 0.1 = 0.4 (i=4)
0.20 m MgCl2
im = 3 x 0.2 = 0.6 (i=3)
0.35 m NaCl,
im = 2 x 0.35 = 0.7 (i=2)
Thus, the order of decreasing freezing point is -
0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,
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Write a balanced equation for the dissolution of sodium carbonate (Na2CO3) in water. Find the number of moles of Na+ produced when 0.207 mol of sodium carbonate dissolves.
Answer:
in the presence of water H2O
Na2CO3 (S) --> 2Na+ (aq)+ (CO3)2-(aq)
One mole of sodium carbonate produces two moles of Na+ ions
Therefore 0.207 moles produces 0.414 moles of Na+ ions
= 0.414 moles of Na+ ions
Explanation:
In water
Na2CO3 --> 2Na+ (aq)+ (CO3)2-(aq)
In a limited reaction, the carbonate ion reacts with the water molecules as follows
(CO3)2-(aq) + H2O←→HCO3-(aq) + OH-(aq)
sodium carbonate or soda ash dissolves in water to give 2 sodium cations and one carbonate anion
The balanced equation for the dissolution of sodium carbonate in water produces two moles of Na+ for every mole of Na2CO3. For 0.207 mol of sodium carbonate, it yields 0.414 mol of Na+ ions.
Explanation:The balanced equation for the dissolution of sodium carbonate (Na2CO3) in water is:
Na2CO3(s) → 2Na+(aq) + CO3^2-(aq)
When 0.207 mol of sodium carbonate dissolves in water, it produces two moles of Na+ ions for every mole of sodium carbonate. Therefore, to find the number of moles of Na+ produced we multiply:
0.207 mol Na2CO3 × 2 mol Na+ / 1 mol Na2CO3 = 0.414 mol Na+
A central component of nonmaterial culture is __________, which are the mental acceptance or conviction that certain things are true or real.
Answer: Beliefs
Explanation: This is a fundamental definition in sociology; the study of society, human social interaction and the rules and processes that bind and separate people not only as individuals, but as members of associations, groups and institutions.
Beliefs are ordinarily defined as the mental acceptance or conviction that certain things are true, real or just the way they are.
Beliefs are a central component of nonmaterial culture, influencing behaviors, values, and norms in a society.
Explanation:A central component of nonmaterial culture is beliefs, which are the mental acceptance or conviction that certain things are true or real. These beliefs can vary widely from one culture to another and can shape behaviors, values, and norms within a society. For example, in some cultures, the belief in an afterlife may lead to specific burial rituals, while in other cultures, the belief in a strict diet may shape what foods are considered acceptable or taboo.
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Beliefs are a central component of nonmaterial culture, representing the mental acceptance or conviction that certain things are true or real. They set the foundation for individuals' perception of the world and guide their behaviors.
Explanation:A central component of nonmaterial culture is beliefs, which are the mental acceptance or conviction that certain things are true or real. Nonmaterial culture refers to the intangible aspects of a society, including its values, norms, and ideologies. These beliefs form the foundation of how individuals perceive and interact with the world, guiding behaviors and influencing societal standards. For example, the belief in equality may cultivate a culture of fairness and non-discrimination in a society.
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A helium balloon containing 0.100 mol of gas occupies a volume of 2.4 L at 25 C and 1.0 atm. how many moles have we added if we inflate it to 5.6 L?
Answer: 0.13mol
Explanation:Please see attachment for explanation
What is the molarity (M) of chloride ions in a solution prepared by mixing 155 ml of 0.276 M calcium chloride with 384 ml of 0.471 M aluminum chloride?
Answer: The concentration of [tex]Cl^-[/tex] ions in the resulting solution is 1.16 M.
Explanation:
To calculate the molarity of the solution after mixing 2 solutions, we use the equation:
[tex]M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of the [tex]CaCl_2[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of the [tex]AlCl_3[/tex]
We are given:
[tex]n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL[/tex]
Putting all the values in above equation, we get
[tex]M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M[/tex]
The concentration of [tex]Cl^-[/tex] ions in the resulting solution will be same as the molarity of solution which is 1.16 M.
Hence, the concentration of [tex]Cl^-[/tex] ions in the resulting solution is 1.16 M.
The molarity of Cl¯ in the solution made by mixing 155 mL of 0.276 M CaCl₂ with 384 mL of 0.471 M AlCl₃ is 1.165 M
We'll begin by calculating the number of mole of chloride ion, Cl¯ in each solution.
For CaCl₂:Volume = 155 mL = 155 / 1000 = 0.155 L
Molarity = 0.276 M
Mole of CaCl₂ =?Mole = Molarity x Volume
Mole of CaCl₂ = 0.276 × 0.155
Mole of CaCl₂ = 0.04278 moleCaCl₂(aq) —> Ca²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole of CaCl₂ contains 2 moles of Cl¯
Therefore,
0.04278 mole of CaCl₂ will contain = 0.04278 × 2 = 0.08556 mole of Cl¯
Thus, 0.08556 mole of Cl¯ is present in 155 mL of 0.276 M of CaCl₂.
For AlCl₃:Volume = 384 mL = 384 / 1000 = 0.384 L
Molarity = 0.471 M
Mole of AlCl₃ =?Mole = Molarity x Volume
Mole of AlCl₃ = 0.471 × 0.384
Mole of AlCl₃ = 0.180864 moleAlCl₃(aq) —> Al³⁺(aq) + 3Cl¯(aq)
From the balanced equation above,
1 mole of AlCl₃ contains 3 moles of Cl¯
Therefore,
0.180864 mole of AlCl₃ will contain = 0.180864 × 3 = 0.542592 mole of Cl¯
Thus, 0.542592 mole of Cl¯ is present in 384 ml of 0.471 M of AlCl₃
Next, we shall determine the total mole of Cl¯ in the resulting solution.Mole of Cl¯ in CaCl₂ = 0.08556 mole
Mole of Cl¯ in AlCl₃ = 0.542592
Total mole = 0.08556 + 0.542592
Total mole = 0.628152 mole Next, we shall determine the total volume of the resulting solutionVolume of CaCl₂ = 0.155 L
Volume of AlCl₃ = 0.384 L
Total volume = 0.155 + 0.384
Total volume = 0.539 LFinally, we shall determine the molarity of Cl¯ in the resulting solutionTotal mole = 0.628152 mole
Total volume = 0.539 L
Molarity of Cl¯ =?Molarity = mole / Volume
Molarity of Cl¯ = 0.628152 / 0.539
Molarity of Cl¯ = 1.165 MTherefore, the molarity of Cl¯ in the resulting solution is 1.165 M
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Predict the mass of oxygen required to react with 14 g of nitrogen to make N2O5 if 16 g of oxygen reacts with 14 g of nitrogen to make N2O2.
The mass of oxygen reacted with 14 g nitrogen has been 16 grams.
The formation of dinitrogen dioxide has been mediated with the reaction of oxygen with nitrogen.
Mass of Oxygen requiredThe mass of oxygen required for the reaction has been given with the balanced chemical equation as:
[tex]\rm N_2\;+\;O_2\;\rightarrow\;N_2O_2[/tex]
Thus, for the reaction with 1 mole of oxygen, 1 mole of nitrogen has been required.
The mass of 1 mole oxygen has been 32 g.
The mass of 1 mole nitrogen has been 28 g.
Thus, for the reaction with 14 g nitrogen, the mass of oxygen required has been:
[tex]\rm 28\g\;N_2=32\;g\;O_2\\14\;g\;N_2=\dfrac{32}{28}\;\times\;14\g\;O_2\\14\;g\;N_2=165\;g\;O_2[/tex]
The mass of oxygen reacted with 14 g nitrogen has been 16 grams.
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The mass of oxygen required to react with 14 g of nitrogen to make [tex]N_2O_5[/tex] is 80 g.
The balanced chemical equation for the formation of [tex]N_2O_5[/tex] from nitrogen [tex](N_2)[/tex] and oxygen [tex](O_2)[/tex] is:
[tex]\[ N_2 + 5O_2 \rightarrow 2N_2O_5 \][/tex]
The balanced chemical equation for the formation of [tex]N_2O_2[/tex] from nitrogen [tex](N_2)[/tex] and oxygen [tex](O_2)[/tex] is:
[tex]\[ N_2 + 2O_2 \rightarrow 2N_2O_2 \][/tex]
From the second equation, we can see that 16 g of oxygen reacts with 14 g of nitrogen to form [tex]N_2O_2[/tex]. This gives us the molar ratio of oxygen to nitrogen for [tex]N_2O_2[/tex], which is 2:1 by mass.
Now, we need to find out the molar ratio of oxygen to nitrogen for [tex]N_2O_5[/tex]. From the first equation, the molar ratio of oxygen to nitrogen is 5:1.
Since the mass of nitrogen (14 g) is the same in both reactions, we can directly compare the ratios of oxygen used in both reactions.
For [tex]N_2O_2[/tex], we have 16 g of oxygen for every 14 g of nitrogen, and for [tex]N_2O_5[/tex], we need 5 times as much oxygen for the same amount of nitrogen.
Therefore, the mass of oxygen required for [tex]N_2O_5[/tex] is:
[tex]\[ \frac{5}{2} \times 16 \text{ g of oxygen} = 40 \text{ g of oxygen} \][/tex]
Using the molar masses of nitrogen (14 g/mol for [tex]N_2[/tex]) and oxygen (32 g/mol for [tex]O_2[/tex]), we can calculate the amount of oxygen needed:
[tex]\[ \frac{14 \text{ g of N2}}{28 \text{ g/mol of N2}} \times \frac{5 \text{ mol of O2}}{1 \text{ mol of N2}} \times 32 \text{ g/mol of O2} = 80 \text{ g of O2} \][/tex]
A 22.25 g piece of metal is dropped into a graduated cylinder with 126.3 mL of water. If the density of the metal is 4.43 g/cm3, what is the total volume in the graduated cylinder in mL with the correct number of significant digits?
Answer:
131.2 mL would be the total volume in the graduated cylinder.
Explanation:
Let's work with the density of metal to find out its volume
Metal density = Metal mass / metal volume
4.46 g/mL = 22.25 g / metal volume
Metal volume = 22.25 g / 4.46 g/mL = 4.98 mL
If we suppose aditive volume, total volume in the graduated cylinder will be:
126.3mL + 4.9mL = 131.2 mL
A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solution is 95.0%H2SO4 by mass and has a density of 1.84 g/mL.
Part A
Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kg of sulfuric acid solution?
Express your answer with the appropriate units
Answer:
6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.
Explanation:
Mass of sulfuric acid solution = [tex]6.05\times 10^3 kg=6.05\times 10^6 g[/tex]
[tex]1 kg = 10^3 g[/tex]
Percentage mass of sulfuric acid = 95.0%
Mass of sulfuric acid = [tex]\frac{95.0}{100}\times 6.05\times 10^6 g[/tex]
[tex]=5,747,500 g[/tex]
Moles of sulfuric acid = [tex]\frac{5,747,500 g}{98 g/mol}=58,647.96 mol[/tex]
[tex]H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O[/tex]
According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.
Then 58,647.96 moles of sulfuric acisd will be neutralized by :
[tex]\frac{1}{1}\times 58,647.96 mol=58,647.96 mol[/tex] of sodium carbonate
Mass of 58,647.96 moles of sodium carbonate :
[tex]106 g/mol\times 58,647.96 mol=6,216,683.76 g[/tex]
6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg
6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.
To neutralize 6.05×10^3 kg of sulfuric acid solution, we would need approximately 6.21×10^3 kg of sodium carbonate, based on the stoichiometry of the chemical reaction between these two compounds.
Explanation:The problem involves a chemical reaction between sulfuric acid (H2SO4) and sodium carbonate (Na2CO3). The balanced equation for this reaction is: H2SO4 + Na2CO3 -> Na2SO4 + H2O + CO2. From this equation, we can see that 1 mole of H2SO4 reacts with 1 mole of Na2CO3. Thus, the mass of Na2CO3 needed can be obtained by first determining the mass of H2SO4 present. Then, we convert the mass of H2SO4 to moles using its molar mass, and finally use the molar mass of Na2CO3 to find the mass of Na2CO3 needed.
First, we calculate the mass of H2SO4 in the solution: mass_H2SO4 = 0.95 * (6.05×10^3 kg of H2SO4 solution) = 5747.5 kg of H2SO4.
To convert this mass into moles, we divide by the molar mass of H2SO4, which is approximately 98.08 g/mol (or 0.09808 kg/mol). So, moles_H2SO4 = mass_H2SO4 / molar_mass_H2SO4 = 5747.5 kg / 0.09808 kg/mol = 58.6×10^3 mol.
From the balanced chemical equation, we know that 1 mole of H2SO4 neutralizes 1 mole of Na2CO3. Thus, to neutralize all the H2SO4, we need 58.6×10^3 mol of Na2CO3. Multiplying this number of moles by the molar mass of Na2CO3 (approximately 105.988 g/mol or 0.105988 kg/mol), we obtain the mass of Na2CO3 required: mass_Na2CO3 = moles_Na2CO3 * molar_mass_Na2CO3 = 58.6×10^3 mol * 0.105988 kg/mol = 6.21×10^3 kg of Na2CO3.
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780. mm Hg to kPa,
I'm struggling to get the answer with the correct sig figs ;(
Answer:
103.9 kPa
Explanation:
It's more easy to convert mmHg to Pa and then /1000
Let's try the rule of three:
760 mmHg is 101325 Pa
780 mmHg (780 mmHg . 101325 Pa) / 760mmHg = 103991.4 Pa
1 kPa = 1000 Pa
103991.4 Pa / 1000 = 103.9 kPa
Silver has two naturally occurring isotopes with the following isotopic masses: 47Ag 107 – 106.90509 47Ag 109 – 108.9047 The average atomic mass of silver is 107.8682 amu. The abundance of the lighter of the two isotopes is __________. 3/1
A) 0.2422B) 0.4816C) 0.5184D) 0.7578E) 0.9047
Answer:
c) .51835
Explanation:
Let the relative abundance of the lighter of the two isotopes be X we have
Then the relative abundance of the heavier isotope is then (1-X)
Whereby we have that in nature the amount of the lighter silver found in proportion is X and the heavier isotope of silver is present as (1-X) proportion in nature.
To calculate the relative atomic mass of silver, we have
(Mass of light weight silver)×X + (mass of heavier isotope of silver×(1-X) = relative atomic mass of silver
106.90509(X) + 108.9047(1-X)
108.9-108.9(x)+106.9(x) = 107.87
-2x-1.03 = 0.517450902926
Closest answer is c
c) .5184
The relative atomic mass of isotopes is the weighted average by the mole-fraction of abundance of these isotopes which gives the atomic weight that is listed for that element on the periodic table.
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹ : ---> 2SO₃g + 2SO₂g O₂g Suppose a vessel contains SO₃ at a concentration of 1.44M . Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.
Answer : The concentration of SO₃ in the vessel 0.240 seconds later is, 0.24 M
Explanation :
The expression used for second order kinetics is:
[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]
where,
k = rate constant = [tex]14.1M^{-1}s^{-1}[/tex]
t = time = 0.240 s
[tex][A_t][/tex] = final concentration = ?
[tex][A_o][/tex] = initial concentration = 1.44 M
Now put all the given values in the above expression, we get:
[tex]14.1\times 0.240=\frac{1}{[A_t]}-\frac{1}{1.44}[/tex]
[tex][A_]t=0.24M[/tex]
Therefore, the concentration of SO₃ in the vessel 0.240 seconds later is, 0.24 M
For action potential generation and propagation, are there any other cation channels that could substitute for the voltage-gated sodium channels if the sodium channels were blocked?
Final answer:
Other cation channels, such as voltage-gated calcium and potassium channels, can substitute for the function of blocked voltage-gated sodium channels in action potential generation and propagation.
Explanation:
If the voltage-gated sodium channels are blocked, other cation channels can substitute for their function in action potential generation and propagation. One such cation channel is the voltage-gated calcium channel. Although calcium channels primarily play a role in synaptic transmission, they can also contribute to the depolarization phase of the action potential in the absence of functional sodium channels.
Another cation channel that can substitute for voltage-gated sodium channels is the voltage-gated potassium channel. While potassium channels primarily repolarize the membrane during the action potential, they can also contribute to the depolarization phase in the absence of sodium channels.
It is important to note that these cation channels may not fully replicate the function of voltage-gated sodium channels, and the exact impact on action potential generation and propagation can vary depending on the specific circumstances.
What is the coefficient of Ca(OH)2 in the equation Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O when balanced using the smallest possible coefficients?
Answer:
The coefficient of Ca(OH)2 is 1
Explanation:
Step 1: unbalanced equation
Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O
Step 2: Balancing the equation
On the right side we have 2x N (in Ca(NO3)2 ) and 1x N on the left side (in HNO3). To balance the amount of N on both sides, we have to multiply HNO3 by 2.
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + H2O
On the left side we have 4x H (2xH in Ca(OH)2 and 2x H in HNO3), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side, by 2.
Now the equationis balanced.
Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O
The coefficient of Ca(OH)2 is 1
Final answer:
The coefficient of Ca(OH)2 is 1.
Explanation:
The coefficient of Ca(OH)2 in the equation Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O when balanced using the smallest possible coefficients is 1.
An unknown compound contains only C, H, and O. Combustion of 6.10 g of this compound produced 14.9 g of CO₂ and 6.10 g of H₂O. What is the empirical formula of the unknown compound?
Answer:
The answer to your question is C₃H₃O
Explanation:
Data
Combustion of a compound C, H, O
mass = 6.10 g
mass CO2 = 14.9 g
mass of water = 6.10 g
Reaction
Cx Hy Oz + O2 ⇒ CO2 + H2O
Process
1.- Calculate the moles of carbon
44 g of CO2 -------------- 12 g of carbon
14.9 g of CO2 ------------- x
x = (14.9 x 12) / 44
x = 4.06 g
12 g of C ------------------ 1 mol
4.06 g ------------------- x
x = (4.06 x 1) / 12
x = 0.34 moles
2.- Calculate the moles of hydrogen
18 g of water ------------- 1 g of hydrogen
6.10 g of water ---------- x
x = (6.10 x 1) / 18
x = 0.33 g
1 g of H ---------------- 1 mol of H
0.33 g ---------------- x
x = (0.33 x 1) / 1
x = 0.33 moles of H
3.- Calculate the mass of oxygen
mass of Oxygen = 6.10 - 4.06 - 0.33
= 1.71 g
16 g of O --------------- 1 mol of O
1.71 g of O ------------- x
x = (1.71 x 1) / 16
x = 0.11 moles
4.- Divide by the lowest number of moles
Carbon = 0.34 / 0.11 = 3
Hydrogen = 0.33 / 0.11 = 3
Oxygen = 0.11 /0.11 = 1
5.- Write the empirical formula
C₃H₃O
Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water. What mass of hydrochloric acid will be needed to produce 5.00 x 10 3 kg of calcium chloride?
Answer:
3284.3 kg is the mass of HCl needed to produce 5000 kg of CaCl₂
Explanation:
This is the reaction:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Let's convert the mass of the product to moles (mass (g) / molar mas)
Firstly we must convert kg to g
1 kg = 1000 g
5×10³ kg . 1000 g / 1 kg = 5×10⁶ g
5×10⁶ g / 110.98 g/mol = 45053.1 moles
Ratio is 1:2 So, let's make a rule of three:
1 mol of CaCl₂ was produced by the use of 2 moles of HCl
So 45053.1 moles of salt, were produced by, the double of moles
45053.1 mol . 2 → 90106.2 moles
Let's convert the moles, to mass (mol . molar mass)
90106.2 mol . 36.45 g/mol = 3284375 g → 3284.3 kg
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.4-L bulb, then filled it with the gas at 1.00 atm and 22.0 ∘C and weighed it again. The difference in mass was 5.1 g . Identify the gas.
Answer : The diatomic gas is nitrogen gas, N₂.
Explanation :
First we have to calculate the moles of gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of gas = 1.00 atm
V = Volume of gas = 4.4 L
n = number of moles of gas = ?
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of gas = [tex]22.0^oC=273+22.0=295.0K[/tex]
Putting values in above equation, we get:
[tex]1.00atm\times 4.4L=n\times (0.0821L.atm/mol.K)\times 295.0K[/tex]
[tex]n=0.1817mol[/tex]
Now we have to calculate the molar mass of gas.
[tex]\text{Molar mass of gas}=\frac{\text{Given mass of gas}}{\text{Moles of gas}}[/tex]
[tex]\text{Molar mass of gas}=\frac{5.1g}{0.1817mol}=28.07g/mol[/tex]
As we are given that the gas is diatomic X₂.
As, 2 atoms of gas X has mass = 28.07 g/mol
So, 1 atom of gas will have mass = [tex]\frac{28.07}{2}=14.04g/mol[/tex]
From this we conclude that the nitrogen atom has mass of 14.04 g/mol.
Thus, the diatomic gas is nitrogen gas, N₂.
A compound analyzes to be 37.5% C, 12.5% H, and 50.0% O. The molecular mass of the compound is 93.0. What is the molecular formula of the compound? (Type your answer using the format CH4 for CH4. Keep the elements in the order given.)
Answer: The molecular formula of the compound is [tex]C_3H_{12}O_3[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 37.5 g
Mass of H = 12.5 g
Mass of O = 50.0 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{37.5g}{12g/mole}=3.125moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{12.5g}{1g/mole}=12.5moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{50.0g}{16g/mole}=3.125moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =[tex]\frac{3.125}{3.125}=1[/tex]
For H= [tex]\frac{12.5}{3.125}=4[/tex]
For O =[tex]\frac{3.125}{3.125}=1[/tex]
The ratio of C: H: O = 1: 4: 1
Hence the empirical formula is [tex]CH_4O[/tex].
The empirical weight of [tex]CH_4O[/tex] = 1(12)+4(1)+1(16)= 32g.
The molecular weight = 93.0 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{93}{32}=3[/tex]
The molecular formula will be=[tex]3\times CH_4O=C_3H_{12}O_3[/tex]
The molecular formula of the compound is C3H12O3. This was determined by converting percent composition to grams, then to moles to find a simplest ratio for the empirical formula and comparing its mass with the molecular mass.
Explanation:The compound could firstly be analyzed into its empirical formula by treating the percentage composition directly as masses, which would then provide us the simplest ratio of elements. Hence for 37.5g of C we have 37.5/12.01 = ~3.1 moles of C, for 12.5g of H we have 12.5/1.01 = ~12.4 moles of H, and for 50.0g of O we have 50.0/16.00 = ~3.1 moles of O. The simplest ratio (empirical formula) will then be C3H12O3, with empirical formula mass equals to 94.1 g/mol. This mass is very close to the mentioned molecular mass of 93.0 g/mol, so our molecular formula is presumably the same as the empirical formula. Therefore, the molecular formula of the compound is C3H12O3.
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HURRY PLEASE HELP!!!!!!!!
Which of the following best explains why the bonds in ammonia (NH3) are polar covalent?
A. Nitrogen has a greater number of isotopes than hydrogen.
B. Nitrogen has a greater electronegativity value than hydrogen.
C. Nitrogen has a mass number that is 14 times that of hydrogen.
D. Nitrogen has seven neutrons, but hydrogen does not have any neutrons.
Answer:
Option B
Explanation:
As nitrogen is more electronegative than hydrogen atom, it attracts the electronic cloud towards it center. This results in poles formation that is partial positive charge on hydrogen and partial negative charge on hydrogen atom. Hence, it makes polar polar covalent bond.
The statement which best explain why the bonds in ammonia is polar covalent is " nitrogen has greater electronegativity than hydrogen. Thus option C is correct.
What is electronegativity?Electronegativity is ability of an atom to attracts bonded pair electrons towards it. The atoms which can acquire stability by accepting one ore more electrons are electronegative.
Nitrogen is an electronegative atom and its valency is 3 means it needs 3 more electron to attain stability. Thus nitrogen is able to attracts the shared electron from hydrogen and forms a partial negative charge.
Hydrogen is an electropositive atom and it forms a partial positive charge by releasing the shared electron towards nitrogen. This charge separation creates there a polarity. That's the bond type is called polar covalent.
Hence, the bond type in ammonia is polar covalent due to the higher electronegativity of nitrogen than hydrogen.
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Yeast and other organisms can convert glucose (C6H12O6) to Ethanol (CH3CH2OH) through a process called Alcoholic Fermentation. The Net reaction is:
C6H12O6------2C2H5OH(l) +2CO2(g)
Calculate the mass of glucose required to produce 2.25L of CO2 measured at P=1 atm and T=295k
Answer:
The answer to your question is 8.28 g of glucose
Explanation:
Data
Glucose (C₆H₁₂O₆) = ?
Ethanol (CH₃CH₂OH)
Carbon dioxide (CO₂) = 2.25 l
Pressure = 1 atm
T = 295°K
Reaction
C₆H₁₂O₆ ⇒ 2C₂H₅OH(l) +2CO₂(g)
- Calculate the number of moles
PV = nRT
Solve for n
[tex]n = \frac{PV}{RT}[/tex]
Substitution
[tex]n = \frac{(1)(2.25)}{(0.082)(295)}[/tex]
Simplification
n = 0.092
- Calculate the mass of glucose
1 mol of glucose --------------- 2 moles of carbon dioxide
x --------------- 0.092 moles
x = (0.092 x 1) / 2
x = 0.046 moles of glucose
Molecular weight of glucose = 180 g
180 g of glucose --------------- 1 mol
x g ---------------0.046 moles
x = (0.046 x 180) / 1
x = 8.28 g of glucose
In the paleoatmosphere, the primary gases were _____. a) oxygen, carbon dioxide, nitrogen b) carbon dioxide, nitrogen, and water vapor c) oxygen and water vapor d) carbon dioxide and oxygen
Answer: Option A. oxygen, carbon dioxide, nitrogen
Explanation:
Can you determine the average atomic mass for an element if you know only the atomic number and mass number of an atom? If not, what other information would you need?
Answer: To determine the average atomic mass, we require abundance of every isotope in the nature.
Explanation:
Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
For Example: There are 3 isotopes of silicon, Si-28, Si-29 and Si-30.
We know that:
Average atomic mass of silicon = 28.086 amu
In order to calculate the average atomic mass of an element, we need the atomic mass of every isotope and also the abundance of every isotope in the nature.
Percentage abundance of Si-28 isotope = 92.2 %
Percentage abundance of Si-29 isotope = 4.7 %
Percentage abundance of Si-30 isotope = 3.1 %
Hence, to determine the average atomic mass, we require abundance of every isotope in the nature.
Describe a sigma bond.A) overlap of two f orbitalsB) end to end overlap of p orbitalsC) s orbital overlapping with the side of a p orbitalD) side by side overlap of p orbitalsE) p orbital overlapping with a d orbital
Answer:
B - End to end overlap of p overlap
Explanation:
Sigma bonds (σ bonds) are the strongest type of covalent chemical bondSigma bonds are formed by end-to-end overlapping. Sigma bonds can occur between any kind of atomic orbitals;The combination of overlapping orbitals can be s-s, s-pz or pz-pz.
Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculate the standard enthalpy of formation of acetone.
Complete combustion of 1 mol of acetone (C3H6O) liberates 1790 kJ:
C3H6O(l)+4O2(g)?3CO2(g)+3H2O(l)?H?=?1790kJ
Answer:
ΔHacetone = - 247.5 kJ/mol
Explanation:
The enthalpy equation is as follows
ΣnΔHproducts – ΣmΔHreactants =ΔHreaction
3×ΣnΔH(CO2(g)) + 3×ΣnΔH(H2O) - ΣnΔH (C3H6O(l)) =ΔHreaction = -1790 kJ/mol
[3(-393.5) + 3(-285.8)] – ΔHacetone
= -1790 kJ/mol
(-1180.5 – 857.4)kJ/mol - ΔHacetone =
-1790 kJ/mol
-2037.9 kJ/mol - ΔHacetone
= -1790kJ/mol
-2037.9 kJ/mol + 1790kJ/mol = ΔHacetone
- 247.5 kJ/mol = ΔHacetone
ΔHacetone = - 247.5 kJ/mol
To determine the standard enthalpy of formation of acetone (C3H6O), we can use the combustion reaction equation and the standard enthalpies of formation of other compounds involved. By applying Hess's Law and the enthalpy change of the reaction, we can calculate the standard enthalpy of formation.
Explanation:The standard enthalpy of formation of acetone (C3H6O) can be calculated using the given equation for the combustion of acetone and the standard enthalpies of formation of O2(g), CO2(g), and H2O(l). For the combustion reaction:
C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(l)
The enthalpy change or ΔH of the reaction is -1790 kJ. Using Hess's Law and the enthalpy change, we can determine the standard enthalpy of formation of acetone.
We can set up an equation using the standard enthalpy of formation values:
ΔH = Σ(ΔHf(products)) - Σ(ΔHf(reactants))
By rearranging the equation, we can solve for the standard enthalpy of formation of acetone.
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If a particular ore contains 55.1 % calcium phosphate, what minimum mass of the ore must be processed to
obtain 1.00 kg of phosphorus?
Express your answer with the appropriate units.
Answer:
6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.
Explanation:
Given
An ore has 51% Calcium phosphate
To find
The minimum mass of ore to be processed to get 1.00 Kg of phosphorous
First find the mass of phosphorous in 1 mole = molar mass of Calcium phosphate, Ca₃(PO₄)₃.
Molar mass of Ca₃(PO₄)₃ is 310 g
Molar mass of P is 31 g
1 mole of Ca₃(PO₄)₃ has 3 atoms of phosphorous
i.e 310 g of Ca₃(PO₄)₃ has 3× 31g of P
= 93 g P
93 g P wil be present in 310 g of Ca₃(PO₄)₃
1 Kg or 1000 g of P will be in (1000÷93) ×310
=3333.33 g of Ca₃(PO₄)₃
but the ore has only 55.1% Ca₃(PO₄)₃
i.e
100 g of Ca₃(PO₄)₃ will have 55.1g Ca₃(PO₄)₃
we need 3333.33g of Ca₃(PO₄)₃
100 g of ore will have 55.1g Ca₃(PO₄)₃
3333.33g of Ca₃(PO₄)₃ will be present in
(3333.33÷ 55.1) ×100
= 6049.60 g of the ore
So 6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.
Which of the following is the best example of an isolated system?
A. A bus at a bus station
B. An astronaut in a spaceship
C. A motorcycle in a garage
D. A car on the freeway
An astronaut in a spaceship. Hence, option B is correct.
What is an isolated system?A system that does not interact with its surroundings.
To stay physiologically, emotionally and psychologically fit in isolation, astronauts exercise, eat healthily, follow a sleep and work schedule, and make time for leisure activities.
Hence, an astronaut in a spaceship is an isolated system.
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Based on extrapolation of your lab results, write the dissolution reaction for the ionic solids below. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)(A) Mn(ClO4)2(B) Li2CO3The two solutions, when mixed, will have two cations and two anions.(C) Based on your lab results, enter the cation and anion for which a precipitate will form. (Separate substances in a list with a comma.)(D) Write the net precipitation reaction that occurs. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
Answer:
The answer has been provided in the attachment
Explanation:
The step by step explanation is as given in the attachment below.
The dissolution reactions for Mn(ClO4)2 and Li2CO3 are Mn(ClO4)2(s) -> Mn2+(aq) + 2ClO4-(aq) and Li2CO3(s) -> 2Li+(aq) + CO3^2-(aq) respectively. The cations Mn^2+ and Li+ will react with the anions ClO4^- and CO3^2- respectively to form a precipitate. The net precipitation reaction is Mn2+(aq) + 2ClO4-(aq) + 2Li+(aq) + CO3^2-(aq) -> Mn(ClO4)2(s) + Li2CO3(s).
Explanation:(A) The dissolution reaction for Mn(ClO4)2 is:
Mn(ClO4)2(s) → Mn2+(aq) + 2ClO4-(aq)
(B) The dissolution reaction for Li2CO3 is:
Li2CO3(s) → 2Li+(aq) + CO32-(aq)
(C) Based on the lab results and the solutions, a precipitate will form when the cations Mn2+ and Li+ react with the anions ClO4- and CO32- respectively.
(D) The net precipitation reaction that occurs is:
Mn2+(aq) + 2ClO4-(aq) + 2Li+(aq) + CO32-(aq) → Mn(ClO4)2(s) + Li2CO3(s)
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If you want to use a serial dilution to make a 1/50 dilution. The first dilution you make is a 1/5 dilution with a total volume of 1 ml. To make that dilution you would add ______of your stock solution and______ of your solvent. Then you make a 1/10 dilution and add ______ of your first dilution and ______ of your solvent.Select one:a. 0.1 ml, 0.9 ml, 0.2 ml, 0.8 mlb. 0.3 ml, 0.7 ml, 0.1 ml, 0.9 mlc. 0.2 ml, 0.8 ml, 0.1 ml, 0.9 mld. 0.2 ml, 0.8 ml, 0.3 ml, 0.7 mle. None of these answer choice
Answer:
c.
Explanation:
A serial dilution is a dilution that is made fractionated. The stock solution is diluted, then this now solution is diluted, and then successively. The final dilution is the multiplication of the steps dilutions.
The representation of the dilution is v/v (volume per volume) indicates how much of the stock solution is in the total volume of the solution. So 1/5 indicates 1 mL to 5 mL of the solution. If the final volume must be 1 mL, then the stock solution must be 0.2 mL (0.2/1 = 1/5), and the volume of the solvent is 1 mL - 0.2 = 0.8 mL.
The second solution is done with a dilution of 1/10 or 1 mL of the first solution in 10 mL of the total solution. Because the solution has 1 mL, then the volume of the first solution must be 0.1 mL (0.1/1 = 1/10), and the volume of the solvent that must be added is 1 mL - 0.1 mL = 0.9 mL.
the mass of potassium in a one cup serving of this cereal is determined to be 172 mg show a numerical setup for calculating the percent error for the mass of potassium to this serving.
Answer:
[(172 - 170)mg/170 mg]*100% = 2*100%/170 = 1.18%
Explanation:
Generally percent error is calculated by dividing the error by the actual value of the variable at standard conditions. In the problem above, the acceptable or actual mass of potassium in a cup of the cereal is 170 mg and the estimated mass of potassium in a cup of the cereal is 172 mg. Therefore, the percent error = [(172 - 170)mg/170 mg]*100% = 2*100%/170 = 1.18%
Nitrogen (N) is much more electronegative than hydrogen (H). Which of the following statements is correct about the atoms in ammonia (NH3)?
A) Each hydrogen atom has a partial positive charge.
B) The nitrogen atom has a strong positive charge.
C) Each hydrogen atom has a slight negative charge.
D) The nitrogen atom has a partial positive charge.
E) There are covalent bonds between the hydrogen atoms.
In the molecule ammonia (NH3), each hydrogen atom has a partial positive charge and the nitrogen atom has a partial negative charge. There are covalent bonds between the hydrogen atoms.
Explanation:In the molecule ammonia (NH3), each hydrogen atom has a partial positive charge (A) because nitrogen is more electronegative than hydrogen. The nitrogen atom, on the other hand, has a partial negative charge (D) due to the uneven distribution of electrons. The presence of covalent bonds between nitrogen and hydrogen atoms (E) results in the formation of ammonia.
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