HELP ME PLZ AND THANKS I WILL MARK YOU AS BRAINLIEST!!!
Answer:
See explanation.
Explanation:
Hello there!
In this case, since this problem is about gas laws, more specifically about the Gay-Lussac's one since the volume is said to be constant, we can use the following equation for its solution for the final pressure, P2:
[tex]\frac{P_2}{T_2} = \frac{P_1}{T_1}[/tex]
[tex]P_2= \frac{P_1T_2}{T_1}\\\\P_2 =\frac{12.0atm*450K}{300K}\\\\P_2= 18.0atm[/tex]
Thus, we fill in the table as follows:
Initial Final
Pressure 12.0 atm 18.0 atm
Volume 4.0 L 4.0 L
Temperature 300K 450K
Regards!
you have 4.600x 10^1 ml of a kcl solution which has been made up in 6.0000x10^-1 g/ml solution.you are asked to determine the %v/v/v of the kcl solution.
Answer: The %v/v of the given KCl solution is 7.6%.
Explanation:
Given: Volume of solute = [tex]4.6 \times 10^{1} ml[/tex]
Volume of solution = [tex]6.0 \times 10^{-1} g/ml[/tex]
Formula used to calculate %v/v is as follows.
[tex]\frac{volume of solute}{volume of solution} \times 100[/tex]
Substitute the values into above formula as follows.
[tex]\frac{volume of solute}{volume of solution} \times 100\\\frac{4.6 \times 10^{1}}{6.0 \times 10^{-1}} \times 100\\= 7.6[/tex]
Thus. we can conclude that the %v/v of the given KCl solution is 7.6%.
At a given temperature, K = 1.3x10^-2 for the reaction:
N2(g) + 3H2(g) ⇌ 2NH3(g)
Calculate values of K for the following reactions at this temperature.
a. 1/2N2 + 3/2H2(g) ⇌ NH3(g)
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
Answer:
a) 0.11
b)76.9
c) 8.8
d) 1.7*10^-4
Explanation:
Step 1: Data given
K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Formula of K
aA(g) + bB(g) ⇌ cC(g) + dD(g)
K = [C]^c *[D]^d / [A]^a * [B]^b
K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]
Step 3:
a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
1/2N2 + 3/2H2(g) ⇌ NH3(g) =>K' = [tex]\sqrt{K}[/tex]
K' = [tex]\sqrt{1.3*10^-2}[/tex] = 0.11
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
2NH3(g) ⇌ N2(g) + 3H2(g) =>K' = 1/K
K' = 1/(1.3*10^-2) = 76.9
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
=>K' = [tex]\frac{1}{\sqrt{K} }[/tex]
K' = [tex]\frac{1}{\sqrt{1.3*10^-2} }[/tex]
K' = 8.8
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
2N2(g) + 6H2(g) ⇌ 4NH3(g)
K' = K²
K' = (1.3*10^-2)²
K' = 1.7 *10 ^-4
Values of equilibrium constant at given temperature for the following reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
How we calculate equilibrium constant?Equilibrium constant is define as the ration of the concentrations of product to the concentrations of reactant with respect to the exponent of their coefficients.
Given chemical reaction is:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Equilibrium constant for this reaction is:
K = [NH₃]² / [N₂][H₂]³
K = 1.3 × 10⁻² (given)
Equilibrium constant K₁ for below reaction will be written as:1/2N₂(g) + 3/2H₂(g) ⇌ NH₃(g)
K₁ = √K
Because concentration of all given species is 1/2 of the given reaction, so value of K₁ will be written as:
K₁ = √(1.3 × 10⁻²) = 0.11
2NH₃(g) ⇌ N₂(g) + 3H₂(g)
K₂ = 1/K
Because concentration of reactant and products are reciprocal from the concentration of original given reaction, so value of K₂ will be written as:
K₂ = 1/1.3 × 10⁻² = 76.9
NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)
K₃ = 1/√K
Because concentrations of given species is reciprocal as well as half of the given original reaction, so value of K₃ will be written as:
K₃ = 1/√(1.3 × 10⁻²) = 8.8
2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)
K₄ = K²
Because concentrations of given species is double of the given original reaction, so value of K₄ will be written as:
K₄ = (1.3 × 10⁻²)² = 1.7 × 10⁻⁴
Hence, the value of K for given reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
To know more about equilibrium constant, visit the below link:
https://brainly.com/question/12858312
study the reaction given below in which excess magnesium ribbon (Mg)reacts with 50cm of a diluted sulphuric acid solution at room temperature
Questions
what Changes can be made to the following substance to increase the rate of reaction?
5.1.1 Magnesium
5.1.2 Sulphuric acid
Answer:
Magnesium reacts with dilute hydrochloric acid in a conical flask which is ... One student can add the magnesium ribbon to the acid and stopper the flask, ... 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.
Can someone help me answer this please
Answer:
A) 1.3 × 10⁻⁵ mol/L
Explanation:
Step 1: Write the balanced equation for the solution of AgCl
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
Step 2: Make an ICE Chart
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
I 0 0
C +S +S
E S S
If we replace the solubility (S) in the Ksp expression, we get,
Ksp = [Ag⁺] [Cl⁻] = S × S = S²
S = √Ksp = √1.8 × 10⁻¹⁰ = 1.3 × 10⁻⁵ mol/L
Which of the following is true for the percentage yield of a reaction?
Answer:
It is always less than the theoretical yield.
When an electron moves up to higher energy levels, the atom Choose... a photon of light whereas the atom Choose... a photon of light when an electron drops to a lower energy level. The photons emitted from an atom appear as
Answer:
Explanation:
When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.
this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.
A flexible vessel contains 65.8 L of gas at a pressure of 2.07 atm. Under the conditions of constant temperature and constant number of moles of gas, what is the pressure of the gas (in atm) when the volume of the vessel increased by a factor of 16.00
Answer: Pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.
Explanation:
The formula for ideal gas equation is as follows.
[tex]PV = Nk_{b}T[/tex]
where,
[tex]k_{b}[/tex] = Boltzmann constant
N = number of moles
That can also be written as:
[tex]\frac{PV}{T} = constant[/tex]
As pressure and volume are inversely proportional to each other. So, if one of the state variable is increased then the other one will decrease or vice-versa.
So, if volume of the vessel increased by a factor of 16.00 then it means pressure is decreased by a factor of 16.00
Therefore, final volume is as follows.
[tex]65.8 L \times 16.00\\= 1052.8 L[/tex]
Now, final pressure is as follows.
[tex]\frac{2.07}{16.00}\\= 0.129375 atm[/tex]
Initially the product of pressure and volume is as follows.
[tex]PV = 2.07 \times 65.8\\= 136.206[/tex]
Hence, if volume of the vessel increased by a factor of 16.00 and pressure is decreased by a factor of 16.00 then its product is as follows.
[tex]PV = 0.129375 \times 1052.8\\= 136.206[/tex]
Here, product of pressure and volume remains the same.
Thus, we can conclude that pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.
At 50.0 oC, a reinforced tank contains 675.5 grams of gaseous argon and 465.0 g of gaseous molecular chlorine with a total pressure of 4.00 atm. Calculate the following:
a. How many moles of Ar are in the tank?
b. How many moles of Cl, are in the tank?
c. Total moles of gas in the tank.
d. The mole fraction of Ar.
e. The mole fraction of Cl2.
f. The Partial Pressure of Ar.
g. The Partial Pressure of Cl2.
Answer:
For (a): The moles of Ar is 16.94 moles
For (b): The moles of [tex]Cl_2[/tex] is 16.94 moles
For (c): The total number of moles in a tank is 23.47 moles
For (d): The mole fraction of Ar is 0.722
For (e): The mole fraction of [tex]Cl_2[/tex] is 0.278
For (f): The partial pressure of Ar is 2.888 atm
For (g): The partial pressure of [tex]Cl_2[/tex] is 1.112 atm
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For (a):Given mass of Ar = 675.5 g
Molar mass of Ar = 39.95 g/mol
Plugging values in equation 1:
[tex]\text{Moles of Ar}=\frac{675.5g}{39.95g/mol}=16.91 mol[/tex]
For (b):Given mass of [tex]Cl_2[/tex] = 465.0 g
Molar mass of [tex]Cl_2[/tex] = 70.9 g/mol
Plugging values in equation 1:
[tex]\text{Moles of }Cl_2=\frac{465.0g}{70.9g/mol}=6.56 mol[/tex]
For (c):Total moles of gas in the tank = [16.91 + 6.56] mol = 23.47 mol
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)
where n is the number of moles
For (d):Moles of Ar = 16.94 moles
Total moles of gas in the tank = 23.47 mol
Putting values in equation 2, we get:
[tex]\chi_{Ar}=\frac{16.94}{23.47}\\\\\chi_{Ar}=0.722[/tex]
For (e):Total mole fraction of the system is always 1
Mole fraction of [tex]Cl_2[/tex] = [1 - 0.722] = 0.278
Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture.
The equation for Raoult's law follows:
[tex]p_A=\chi_A\times p_T[/tex] .....(3)
where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture
For (f):We are given:
[tex]\chi_{Ar}=0.722\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Ar}=0.722\times 4.00atm\\\\p_{Ar}=2.888atm[/tex]
For (g):We are given:
[tex]\chi_{Cl_2}=0.278\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Cl_2}=0.278\times 4.00atm\\\\p_{Cl_2}=1.112atm[/tex]
what is the characteristics of tropical air mass
Answer:
Explanation:
Continental tropical air masses are extremely hot and dry. Arctic, Antarctic, and polar air masses are cold. The qualities of arctic air are developed over ice and snow-covered ground. Arctic air is deeply cold, colder than polar air masses.
What is the mass of 2.7 L of water?
Compound A, C8H10, absorbed 3 equivalents of H2 on catalytic hydrogenation over a Pd/C catalyst to give B (C8H16). On ozonolysis, compound A gave, among other things, a ketone which was identified as cyclopentanone. On treatment with NaNH2 in NH3, followed by addition of iodomethane, compound A gave a new hydrocarbon, C (C9H12). Draw the structure of Compound A and Compound B.
Answer:
Compound A, C8H10, absorbed 3 equivalents of H2 on catalytic hydrogenation over a Pd/C catalyst to give B (C8H16). On ozonolysis, compound A gave, among other things, a ketone which was identified as cyclopentanone. On treatment with NaNH2 in NH3, followed by addition of iodomethane, compound A gave a new hydrocarbon, C (C9H12). Draw the structure of Compound A and Compound B.
Explanation:
The degree of unsaturation in the given compound A C8H10 is:
DU= (Cn+1)-Hn/2 -Xn/2 +Nn/2
DU=(8+1)-10/2
=9-5
=4
So, the given compound has either three double bonds and a ring or four double bonds or four rings.
Given,
compound A C8H10 absorbed three equivalents of H2 on catalytic hydrogenation that means compound A has three double bonds or it has one double bond and one triple bond.
The structure of compounds A, B and C along with the entire reaction is shown below:
Anyone knows this? I don’t know this
QUESTION :WHICH OF THE FOLLOWING IS AN EXAMPLE OF A CONTROLLED EXPERIMENT TO TEST THIS?
ANSWER:
D. The temperatures of five breakers of 250 mL of water are varied, and 10 g of sugar is added to each breaker.
Equilibrium constants for gases can be expressed in terms of concentrations, Kc, or in terms of partial pressures, Kp. Which one of the following statements regarding Kc and Kp is correct?
a. Kc and Kp are equal when all stoichiometric coefficients in the balanced reaction equation equal one.
b. Kc and Kp are equal when the conditions are standard (P= 1 atm, T=298 K)
c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.
d. Kc and Kp can never be equal.
e. Kc and Kp have the same values but different units.
Answer:
c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to write the relationship between Kc and Kp in terms of the temperature and the change in the stoichiometric coefficients:
[tex]Kp=Kc(RT)^{\Delta \nu}[/tex]
Thus, we can see that the Kp=Kc just when Δυ=0, which is the sum of the coefficients of products minus that of reactants; and therefore, the correct answer will be c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.
Regards!
A certain first-order reaction is 45.0% complete in 65 s. What are the values of the rate constant and the half-life for this process
Answer:
0.01228s⁻¹ = rate constant
Half-life = 56.4s
Explanation:
The first order reaction follows the equation:
ln[A] = -kt + ln[A]₀
Where [A] is amount of reactant after time t = 45.0%, k is rate constante and [A]₀ initial amount of reactant = 100%
ln[45%] = -k*65s + ln[100%]
-0.7985 = -k*65s
0.01228s⁻¹ = rate constant
Half-life is:
Half-life = ln2 / k
Half-life = 56.4s
Ammonium sulfate (NH4)2SO4 is made by reacting 25.0 L of 3.0 mol/L H2SO4 with 3.1× 103 L of NH3 at a pressure of 0.68 atm and a temperature of 298 K according to the following reaction .
NH3(g) + H2SO4(aq) → (NH4)2SO4 (aq)
How many grams of ammonium sulfate are produced?
Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
Explanation:
For [tex]H_2SO_4[/tex]:Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
Molarity of [tex]H_2SO_4[/tex] = 3.0 M
Volume of solution = 25.0 L
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]
For [tex]NH_3[/tex]:The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where,
P = pressure of the gas = 0.68 atm
V = volume of gas = [tex]3.1\times 10^3L[/tex]
n = number of moles of gas = ? moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the gas = 298 K
Putting values in equation 2, we get:
[tex]0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol[/tex]
For the given chemical equation:
[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]
By stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]
As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent
Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]
Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?
Answer:
0.375mL of solution of nalorphine must be injected
Explanation:
A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:
1.5x10⁻³g * (100mL / 0.40g) =
0.375mL of solution of nalorphine must be injectedThe theoretical yield of zinc oxide in a reaction is 486 g. What is the percent
yield if 399 g is produced?
O A. 122%
O B. 4.93%
C. 82.1%
D. 29.6%
Answer:
the correct answer is c
Explanation:
becuase i had the same question
The value of keq for the following reaction is 0.25
SO2(g) + NO2(g) _ SO3(g) + NO(g)
What is the value of at the same temperature if we multiply the reaction by 2
What size volumetric flask would you use to create a 1.00M solution using 166.00 g of KI?
Answer:
A 1 liter volumetric flask should be used.
Explanation:
First we convert 166.00 g of KI into moles, using its molar mass:
Molar mass of KI = Molar mass of K + Molar mass of I = 166 g/mol
166.00 g ÷ 166 g/mol = 1 mol KIThen we calculate the required volume, using the definition of molarity:
Molarity = moles / litersLiters = moles / molarity
1 mol / 1.00 M = 1 LPlease help me with question 7. Thank you so much.
Answer:
The system is not at equilibrium and the reaction will proceed to the left.
Explanation:
Step 1: Write the balanced equation
H₂(g) + CO₂(g) ⇄ CO(g) + H₂O(g)
Step 2: Calculate the reaction quotient (Q)
The reaction is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.
Q = [CO] × [H₂O] / [H₂] × [CO₂]
Q = 0.610 × 0.695 / 0.425 × 0.500 = 2.00
Since Q ≠ Kc, the reaction is not at equilibrium.
Since Q > Kc, the reaction will proceed to the left.
Question 1 of 10
What happens when a solid becomes a liquid?
Answer:it dissolves and evaporates
Explanation:
17. Which of the following is a device that generates electricity using a chemical reaction?
O A. Fuel cell
B. Battery
C. Charging station
O D. Solar panel
Answer:
Hydrogen and fuel cell technologies power cars, buildings and more. But how ... Test your knowledge with this quiz! ... How do fuel cells generate electricity
Answer:
A
Explanation:
fuel cell is a device that converts the chemical energy from fuel into electricity via a chemical reaction with oxygen or another oxidizing agent. Batteries work in a closed system, while fuel cells require their reactants to be replenished.
Groups on the periodic table also correspond with the number of ?
The question is incomplete, the complete question is;
Groups of the periodic table correspond to elements with a. the same color b. the same atomic number c. similar chemical properties d. similar numbers of neutrons
Answer:
similar chemical properties
Explanation:
In the periodic classification of elements, elements are divided into groups and periods. Elements in the same group of the periodic table have the same number of outermost electrons and share very similar chemical properties.
Elements in the same period have the same number of shells and the same maximum energy level of the outermost electron. Chemical properties carry markedly across a period.
A scientist collects a sample that has 2.00 × 1014 molecules of carbon dioxide gas.How many grams is this, given that the molar mass of CO2 is 44.01 g/mol?
Answer:
1.46 × 10⁻⁸ g
Explanation:
Step 1: Given data
Molecules of CO₂: 2.00 × 10¹⁴ molecules
Step 2: Convert molecules to moles
We need a conversion factor: Avogadro's number. There are 6.02 × 10²³ molecules in 1 mole of molecules.
2.00 × 10¹⁴ molecules × 1 mol/6.02 × 10²³ = 3.32 × 10⁻¹⁰ mol
Step 3: Convert moles to mass
We need a conversion factor: the molar mass. The molar mass of CO₂is 44.01 g/mol.
3.32 × 10⁻¹⁰ mol × 44.01 g/mol = 1.46 × 10⁻⁸ g
What are the lengths of the diagonals of the kite?
The answer ( 13 and 8 )
x²=5²+12²
x²=25+144
x²=169
x=13
x²=5²+6²
x²=25+36
x²=61
x=7.8
x=8
the mixture of base and acid
Answer:
Mixture of a Strong Acid and a Strong Base
On mixing a strong acid and strong base neutralization (pH = 7) takes place. The resulting solution may be an acid or base depending on the Concentration. Say, N1, V1 is the strength and volume of the strong acid and N2, V2 is the strength and volume of the strong base
Explanation:
Most introductory chemistry books will teach that the reaction between an acid and a base is called neutralization, and the products formed are water and a salt
What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87m (subject is science) pls answer fast
Answer:
43.5
Explanation:
Hope that helps
ACTUAL YIELD VS THEORETICAL YIELD?
Actual yield over theoretical yield, then multiply by 100
How many joules of heat energy are required to raise the temperature of 100.0 g of aluminum by 120.0°C? The specific heat of aluminum is 0.897 J/g.°C. 2 3
Answer:
10764 J
Explanation:
Remember the equation for specific heat::
q = mcΔT
q = 100 x 0.897 x 120
q = 10764
