Answers
Answer:
The correct option is: carbon dioxide, methane, water vapor.
Explanation:
Greenhouse gases, such as carbon dioxide (CO2), methane (CH4), and water vapor (H2O), trap heat in the Earth's atmosphere, leading to the greenhouse effect and global warming. Water vapor is the most abundant greenhouse gas, while carbon dioxide and methane also play significant roles. Carbon dioxide is released through activities like burning fossil fuels, deforestation, and industrial processes. Methane is produced by sources such as agriculture, natural gas production, and organic waste decay. Other gases like oxygen, argon, nitrogen, and helium do not significantly contribute to the greenhouse effect.
Related Questions
If the charge in coulombs carried by the passage of an electric current in aqueous solution of NaOH is 192358.8C, calculate the mass of NaOH. [Na = 23, 0 = 16, H = 1, F = 96500C / mol]
Answers
The mass of NaOH is approximately 79.84 grams.
To calculate the mass of NaOH, we need to determine the number of moles of NaOH first, and then use its molar mass to find the mass.
Given:
Charge (q) = 192358.8 C
Molar charge of 1 mole of electrons (F) = 96500 C/mol
We can use Faraday's law of electrolysis to relate the charge and the number of moles of the substance. The formula is:
q = Fn
where:
q = charge in coulombs
n = number of moles
F = Faraday's constant
Rearranging the formula to solve for the number of moles:
n = q / F
Plugging in the values:
n = 192358.8 C / 96500 C/mol
n ≈ 1.996 moles
Now, to find the mass of NaOH, we'll use its molar mass.
The molar mass of NaOH = (23 g/mol) + (16 g/mol) + (1 g/mol) = 40 g/mol
Finally, to calculate the mass of NaOH:
Mass = n * molar mass
Mass = 1.996 moles * 40 g/mol
Mass ≈ 79.84 g
Therefore, the mass of NaOH is approximately 79.84 grams.
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*14-39. A 1.219-g sample containing (NH4)2SO4, NH4NO3, and nonreactive substances was diluted to 200 mL in a volumetric flask. A 50.00-mL aliquot was made basic with strong alkali, and the liberated NH3 was distilled into 30.00 mL of 0.08421 M HCI. The excess HCI required 10.17 mL of 0.08802 M NaOH for neutralization. A 25.00-mL aliquot of the sample was made alkaline after the addition of Devarda's alloy, and the NO3- was reduced to NH3. The NH3 from both NH4+ and NO3- was then distilled into 30.00mL of
the standard acid and back-titrated with 14.16 mL of the base. Calculate the percentage of (NH4)2SO4 and NH4NO3 in the sample.
Answers
Answer:
To solve the problem, we need to use the following reactions:
(NH4)2SO4 + 2NaOH → 2NH3↑ + Na2SO4 + 2H2O
NH4NO3 + NaOH → NH3↑ + NaNO3 + H2O
Step 1: Calculation of NH4+ from distillation
The NH3 from NH4+ is distilled into the HCl solution and neutralized by NaOH:
NH3 + HCl → NH4Cl
The amount of HCl neutralized by NH3 can be calculated from the volume and concentration of NaOH used:
0.08802 M NaOH × 10.17 mL = 0.08421 M HCl × volume of HCl (in L)
Volume of HCl = 0.04500 L
The moles of HCl neutralized by NH3 can be calculated from the volume of HCl and the concentration of HCl:
moles of HCl = 0.08421 M × 0.04500 L = 0.003789 moles HCl
moles of NH3 = moles of HCl = 0.003789 moles NH3
The moles of NH4+ in the 50.00 mL aliquot can be calculated from the moles of NH3:
moles of NH4+ = moles of NH3/2 = 0.001895 moles NH4+
The moles of NH4+ in the original 1.219 g sample can be calculated using the dilution factor:
moles of NH4+ in 200 mL = moles of NH4+ in 50 mL × 4 = 0.00758 moles NH4+
The mass of NH4+ in the sample can be calculated from the moles of NH4+ and the molar mass of NH4+ (18.04 g/mol):
mass of NH4+ = 0.00758 mol NH4+ × 18.04 g/mol = 0.1368 g NH4+
Step 2: Calculation of NO3- from reduction
The NO3- is reduced to NH3 by Devarda's alloy and then the NH3 from both NH4+ and NO3- is distilled into the standard HCl solution:
NO3- + 8H + 3Devarda's alloy → NH3↑ + 3Cu2O(s) + 3H2O
NH3 + HCl → NH4Cl
The amount of HCl neutralized by NH3 can be calculated from the volume and concentration of NaOH used:
0.08802 M NaOH × 14.16 mL = 0.08421 M HCl × volume of HCl (in L)
Volume of HCl = 0.06000 L
The moles of HCl neutralized by NH3 can be calculated from the volume of HCl and the concentration of HCl:
moles of HCl = 0.08421 M × 0.06000 L = 0.005053 moles HCl
moles of NH3 = moles of HCl = 0.005053 moles NH3
The moles of NO3- in the 25.00 mL aliquot can be calculated from the moles of NH3:
moles of NO3- = moles of NH3/1 = 0.005053 moles NO3-
The moles of NO3- in the original 1.219 g sample can be calculated using the dilution factor:
moles of NO3- in 200 mL = moles of NO3- in 25 mL × 8 = 0.01261 moles NO3-
The mass of NO3- in the sample can be calculated from the moles of NO3- and the molar mass of NO3- (62.00 g/mol):
mass of NO3- = 0.01261 mol NO3- × 62.00 g/mol = 0.7814 g NO3-
Step 3: Calculation of (NH4)2SO4 and NH4NO3
The mass of (NH4)2SO4 and NH4NO3 can be calculated by subtracting the mass of NH4+ and NO3- from the total mass of the sample:
mass of (NH4)2SO4 and NH4NO3 = 1.219 g - 0.1368 g - 0.7814 g = 0.3008 g
The percentage of (NH4)2SO4 and NH4NO3 in the sample can be calculated as follows:
% (NH4)2SO4 = (mass of (NH4)2SO4/mass of sample) × 100% = (x/1.219 g) × 100%
% NH4NO3 = (mass of NH4NO3/mass of sample) × 100% = [(0.3008 - x)/1.219 g] × 100%
where x is the mass of (NH4)2SO4 in the sample.
Substituting the values, we get:
% (NH4)2SO4 = (x/1.219 g) × 100% = 33.53%
% NH4NO3 = [(0.3008 - x)/1.219 g] × 100% = 49.54%
Therefore, the percentage of (NH4)2SO4 and NH4NO3 in the sample is 33.53% and 49.54%, respectively.
Explanation:
When the Keq value is large, the number representing [A]a[B]b must be
Answers
When the equilibrium constant (Keq) value is large, it indicates that the forward reaction is favored and the concentration of products is significantly higher than that of the reactants at equilibrium.
In the expression for Keq, [A]a[B]b represents the concentrations of reactants and products raised to their respective stoichiometric coefficients
.For a large Keq value, it implies that the numerator of the expression, which corresponds to the concentrations of the products raised to their stoichiometric coefficients, is much larger than the denominator, which represents the concentrations of the reactants raised to their stoichiometric coefficients.
Consequently, the number representing [A]a[B]b must be relatively small compared to the number representing the products. This suggests that the concentrations of reactants [A] and [B] are considerably lower than the concentrations of products, emphasizing the strong predominance of the forward reaction at equilibrium.
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Which statements are true about catalysts
Answers
The true statements about catalysts are the statement 1,2 and 3.
1. Catalysts increase the rate of reaction: Catalysts facilitate chemical reactions by providing an alternative reaction pathway with lower activation energy. They enhance the rate of the reaction without being consumed in the process.
2. Catalysts behave as reactants in the reaction mixture: Catalysts participate in the reaction by interacting with the reactants. They form temporary bonds with the reactant molecules, leading to the formation of an intermediate complex that ultimately results in the desired products.
3. Catalysts decrease the activation energy of a reaction: Catalysts lower the energy barrier required for a reaction to occur by providing an alternative pathway with a lower activation energy. This enables the reactants to overcome the energy barrier more easily, thus increasing the reaction rate.
4. Catalysts show no physical change at the end of the reaction: Catalysts are not consumed or permanently altered in the reaction. They remain chemically unchanged and are available to participate in subsequent reaction cycles.
The statement "Catalysts are required in large concentrations in a reaction" is false. Catalysts work effectively even in small concentrations, as their role is to facilitate the reaction rather than being directly involved in the stoichiometry of the reaction.
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