Answer:
plz mark as BRAINLIEST plz...
Explanation:
●Intellectual factor: The term refers to the individual mental level. ...
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How many grams of sodium chloride are required to make 2.00 L of a solution with a concentration of 0.100 M?
Answer:
Mass = 11.688g
Explanation:
Volume = 2.00L
Molar concentration = 0.100M
Mass = ?
These quantities are relatted by the following equation;
Conc = Number of moles / volume
Number of moles = Conc * Volume = 2 * 0.100 = 0.2 mol
Number of moles = Mass / Molar mass
Mass = Number of moles * Molar mass
Mass = 0.2mol * 58.44g/mol
Mass = 11.688g
Chlorine has an electronegativity value of 3.0. Given the electronegativity of N, O, and P (3.0, 3.5, and 2.1, respectively), which of the following molecules has nonpolar bonds?a. NCl3b. Cl2Oc. PCl3d. All of thesee. None of these
Answer:
a. NCl3
Explanation:
All the elements stated are non metals. Generally the bond between non metals is a covalent bond. However depending on how the electrons are shared, this bond can either be polar or non polar.
A non polar covalent bond is formed when the electrons are shared equally between the atoms.
A way to determine if a bond is non polar or polar covalent is by comparing the electronegativity values.
If the electronegativity difference is less than 0.5, it is regarded as a non polar covalent bond.
Going back to the question;
Between N and Cl; the electronegativity difference is 3.0 - 3.0 = 0
Between Cl and O; the electronegativity difference is 3.5 - 3.0 = 0.5
Between Cl and P; the electronegativity difference is 3.5 - 2.1 = 1.4
From these we can tell that the correct option is option A.
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The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.
Answer:
Half-life at 629K = 252.4min
Explanation:
Using Arrhenius equation:
[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]
And as Half-life in a first order reaction is:
[tex]t_{1/2}=\frac{ln2}{K}[/tex]
We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:
[tex]58.0min=\frac{ln2}{K}[/tex]
K = 0.01195min⁻¹ = K₁
[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]
[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]
[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]
K₂ = 2.75x10⁻³ min⁻¹
And, replacing again in Half-life expression:
[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]
Half-life at 629K = 252.4minThe half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.
The activation energy of a reaction is related to its rate constant as follows:
[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex] (1)
Where:
k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature
We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:
[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex] (2)
Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min
Hence, the rate constant at 652 K is:
[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]
Now, from equation (1) we can find the pre-exponential factor (A):
[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]
With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):
[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]
Finally, the half-life at 629 K is (eq 2):
[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]
Therefore, the half-life at 629 K is 251.1 min.
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One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
Answer:
6.5 mg/L.
Explanation:
Step one: write out and Balance the chemical reaction in the Question above:
NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.
Step two: Calculate or determine the number of moles of AgCl.
So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:
Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.
Step three: Calculate or determine the number of moles of NiCl2.
Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.
Step four: detemine the mass of NiCl2.
Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.
Step five: finally, determine the concentration of NiCl2.
1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.
Without doing any calculations, determine the sign of ΔSsys for each of the following chemical reactions. Drag the appropriate items to their respective bins.
1. 2H30' (aq) + CO23- (aq) - CO2(g) +3H2O(1)
2. CH4(g) + 202,(g) - CO2(g) + 2H2O(l)
3. Mg (s) + Cl2(g) - MgCǐ2(s)
4. SO3(g) + H2O(I) - H2SO4(I)
A. ΔSsys greater than
B. ΔSsys smaller than
Answer:
Answers are in the explanation.
Explanation:
In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:
Entropy of gases >>> entropy of liquid > entropy of solids.
The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.
In the reactions:
1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)
As 1 gas is produced, entropy of products is higher than entropy of reactants. That means ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)
2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.
3. Mg(s) + Cl₂(g) → MgCǐ₂(s)
You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
4. SO₃(g) + H₂O(I) → H₂SO₄(I)
1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:
1. ΔSsys > 0.
2. ΔSsys < 0
3. ΔSsys < 0
4. ΔSsys < 0
Recall:
The states of the reactants and the products in a chemical reaction determines the sign of ΔSsys of the reaction.The entropy of gasses is greater than the entropy of liquid and solids.The entropy of solids is less than the entropy of liquid and gasses.Gasses have the highest entropy, while solids have the least.Thus:
In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.
In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.
In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.
In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.
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Determine whether the following statement about equilibrium is true or false.
(a) When a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants.
(b) When a system is at equilibrium, Keq = 1.
(c) At equilibrium, the rates of the forward reaction and the reverse reaction are equal.
(d) Adding a catalyst to a reaction system will shift the position of equilibrium to the right so there are more products at equilibrium than if there was no catalyst present.
Answer:
(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants
Determining whether the statements about equilibrium is True or False
A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE
B) When a system is at equilibrium, Keq = 1 : TRUE
C) The rates of the forward reaction and the reverse reaction are equal at equilibrium : TRUE
D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE
Reaction at equilibriumIn a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.
A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).
Hence we can conclude that the answers to your questions are as listed above.
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Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
Answer:
[tex]m_{CaCO_3}=0.179gCaCO_3[/tex]
Explanation:
Hello,
In this case, since the undergoing chemical reaction is:
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:
[tex]PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2[/tex]
Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:
[tex]m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3[/tex]
Regards.
The mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g
From the question,
We are to determine the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP.
First, we will determine the number of mole of CO₂ required to be produced
From the formula
PV = nRT
Where
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant
and T is the temperature
Then, we can write that
[tex]n = \frac{PV}{RT}[/tex]
From the question,
V = 40.0 mL = 0.04 L
At STP
P = 1 atm
T = 273.15 K
and
R = 0.08206 L atm mol⁻¹ K⁻¹
Putting the parameters into the formula, we get
[tex]n = \frac{1 \times 0.04}{0.08206 \times 273.15}[/tex]
∴ n = 0.0017845 mole
Now, we will write the balanced chemical equation for the decomposition of CaCO₃
CaCO₃ → CaO + CO₂
This means,
1 mole of CaCO₃ will decompose to produce 1 mole of CO₂
Since 0.0017845 mole of CO₂ is to be produced,
Then,
0.0017845 mole of CaCO₃ would be required
Now, for the mass of CaCO₃ required,
Using the formula
Mass = Number of moles × Molar mass
Molar mass of CaCO₃ = 100.0869 g/mol
∴ Mass of CaCO₃ required = 0.0017845 × 100.0869
Mass of CaCO₃ required = 0.178605 g
Mass of CaCO₃ required ≅ 0.179 g
Hence, the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g
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3. Strontium-90 is produced during the nuclear fission of uranium-235 and is part of nuclear fallout created by weapons testing. If the half-life of Sr-90 is 28 days, how long will it take for grass contaminated with Sr-90 to be safe (<2 percent of the starting radioactivity) for cattle to eat?
A. 158 days
B. 28 days
C. 1 year
D. 158 years
Answer:158 days (D)
Explanation:
It will take 158 days for grass contaminated with Sr-90 to be safe for cattle to eat. Therefore, option (A) is correct.
What is the half-life period?The half-life of a radioactive material is defined as the time that is needed to reduce the initial quantity of a radioactive element to half after disintegration.
The half-life of a radioactive element can be described as the characteristic of the element and does not influence by the initial amount of the radioactive substance.
Given, the half-life of the Strontium-90 = 28 days
The rate constant of the decay can be determined as:
[tex]t_{\frac{1}{2} } =\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{t_{\frac{1}{2} } }[/tex]
k = 0.693/28
k = 0.025 day⁻¹
The concentration of Strontium-90 reduced to less than 2% is safe. Therefore final concentration [A] = 2 % = 0.02
[tex]t = \frac{2.303}{k} log \frac{[A_o]}{[A]}[/tex]
[tex]t = \frac{2.303}{0.02475} log \frac{1}{[0.02]}[/tex]
t = 158 days
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what is the molality of a solution
Answer: The number of moles of a solute per kilogram of solvent
Explanation:
Three structural isomers have the formula C5H12.C5H12. Draw and name the isomers using IUPAC names. Draw the isomer with five carbon atoms in main chain.
Answer:
Explanation:
Answer in attached file .
[tex]pH = -log10[H+ ][/tex]. If [tex][H^+][/tex] is [tex]1.2 * 10^-^6[/tex], what is the [tex]pH[/tex]? (F5 to refresh page if you can't see it)
A laboratory technician drops a 0.0850 kg sample of unknown solid material, at a temperature of 100 oC, into a calorimeter. The calorimeter can, initially at 19.0 oC, is made of 0.150 kg of copper and contains 0.20 kg of water. The final temperature of the calorimeter can, and contents is 26.1 oC. Compute the specific heat of the sample.
Answer:
The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
Explanation:
Given that:
mass of an unknown sample [tex]m_3[/tex] = 0.0850
temperature of the unknown sample [tex]t_{unknown}[/tex] = 100° C
initial temperature of the calorimeter can = 19° C
mass of copper [tex]m_1[/tex] = 0.150 kg
mass of water [tex]m_2[/tex]= 0.20 kg
the final temperature of the calorimeter can = 26.1° C
The objective is to compute the specific heat of the sample.
By applying the principle of conservation of energy
[tex]Q = mc \Delta T[/tex]
where;
[tex]Q_1 +Q_2 +Q_3 = 0[/tex]
i.e
[tex]m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2+m_3 c_3 \Delta T_3 =0[/tex]
the specific heat capacities of water and copper are 4.18 × 10³ J/kg.K and 0.39 × 10³ J/kg.K respectively
the specific heat of the sample [tex]c_3[/tex] can be computed by making [tex]c_3[/tex] the subject of the above formula:
i.e
[tex]c_3 = \dfrac{m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2}{m_3 c_3 \Delta T_3}[/tex]
[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (26.1 -19) + 0.20 \times 4.18 \times 10^3 \times (26.1 -19) }{0.0850 \times (100-26.1 )}[/tex]
[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (7.1) + 0.20 \times 4.18 \times 10^3 \times (7.1) }{0.0850 \times (73.9)}[/tex]
[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]
[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]
[tex]c_3 = \dfrac{6350.95}{6.2815}[/tex]
[tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous compound is left. Further analysis shows that the percentage composition of the anhydrate is 21.90% Ca, 43.14% Se, and 34.97% O.. (Hint: Treat the anhydrous compound and water just as you have treated elements in calculating in the formula of the hydrate.) (Use an asterisk to enter the dot in the formula. If a subscript is 1, omit it.) Find the empirical formula of the anhydrous compound. Find the empirical formula of the hydrate.
Answer:
The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.
Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams
The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,
Moles of water = weight of water/molecular weight
= 42.1 grams / 18 = 2.3
The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,
= 214.2 * 0.2190 = 46.91 grams
The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,
= 214.2 * 0.4314 = 92.40 grams
The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,
= 214.2 * 0.3497 = 74.91 grams
The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17
The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,
92.40/78.96 = 1.17
The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,
74.91/15.999 = 4.68.
Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O
There are eight consitutional isomers with the molecular formula C4H11N.
name and draw a structural formulas for each amine.
Answer:
See figure 1
Explanation:
We have to remember that in the isomer structures we have to change the structure but we have to maintain the same formula, in this case [tex]C_4H_1_1N[/tex].
In the formula, we have 1 nitrogen atom. Therefore we will have as a main functional group the amine group.
In the amines, we have different types of amines. Depending on the number of carbons bonded to the "N" atom. In the primary amines, we have only 1 C-H. In the secondary amines, we have two C-N bonds and in the tertiary amines, we have three C-N bonds.
With this in mind, we can have:
-) Primary amines:
1) n-butyl amine
2) sec-butyl amine including 2 optical isomers
3) isobutyl amine
4) tert-butyl amine
-) Secondary amines:
5) N-methyl n-propyl amine
6) N-methyl isopropyl amine
7) N, N-diethyl amine
-) Tertiary amines:
8) N-ethyl N, N-dimethyl amine
See figure 1
I hope it helps!
The surface temperature on Venus may approach 753 K. What is this temperature in degrees Celsius?
Answer:
461.85 degrees Celsius
Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K sp = 1.4 × 10 –5). 1.5 × 10–2 M 1.4 × 10–5 M 3.0 × 10–2 M 2.4 × 10–2 M None of the above.
Answer:
3.0x10⁻²M
Explanation:
Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:
Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻
When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:
ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:
1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
1.4x10⁻⁵ = [2X]² [X]
1.4x10⁻⁵ = 4X³
3.5x10⁻⁶ = X³
0.015 = X
As [Ag⁺] is 2X:
[Ag⁺] = 0.030 = 3.0x10⁻²M
The answer is:
3.0x10⁻²MA certain electrochemical cell has a cell potential of +0.34 V. Which of the following is a true statement about the electrochemical reaction?
a. The reaction is reactant favored and would be considered an electrolytic cell.
b. The reaction is reactant favored and would be considered a voltaic (galvanic) cell.
c. The reaction is product favored and would be considered an electrolytic cell.
d. The reaction is at equilibrium and is a voltaic (galvanic) cell.
e. The reaction is product favored and would be considered a voltaic (galvanic
Answer:
e. The reaction is product favored and would be considered a voltaic (galvanic) cell
Explanation:
An electrochemical cell produces electrical energy from electrochemical reactions.
A voltaic cell is a type of electrochemical cell that produces electrical energy by spontaneous electrochemical reactions. In a voltaic cell, the cell potential is always positive unlike in an electrolytic cell.
Hence, given the fact that the cell potential is positive, it is a product favoured voltaic cell.
the following glassware was found in a lab drawer: 12 beakers, 10 flasks and 60 test tubes. what percent of the glassware are test tubes?
Answer:
73.1707317073%
=> Approximately 73.2%
Explanation:
Total = 60 + 10 + 12
=> 82
Test tubes are 60/82
=> 30/41
=> 73.1707317073%
=> Approximately 73.2%
Answer:
73.17%
Explanation:
To find the percentage of test tubes to the overall glassware, we need to get the number of test tubes divided by the total number of glassware.
12 beakers + 10 flasks + 60 test tubes = 82 glassware
% test tube = 60 / 82 = .7317 ==> 73.17 %
So 73.17 % of the glassware was test tubes.
Cheers.
What is the value of the equilibrium constant, K, for a reaction for which ∆G° is equal to –5.20 kJ at 50°C?
Answer:
6.93
Explanation:
Step 1: Given data
Standard Gibbs free energy (∆G°): -5.20 kJTemperature (T): 50°CEquilibrium constant (K): ?Step 2: Convert the temperature to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 50°C + 273.15
K = 323 K
Step 3: Calculate K
We will use the following expression.
∆G° = -R × T × ln K
-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K
K = 6.93
Which solution, if either, would create the higher osmotic pressure (compared to pure water): one prepared from 1.0 g of NaCl in 10 mL of water or 1.0 g of CsBr in 10 mL of water
Answer: NaCl would give the higher pressure
Explanation:
Osmotic pressure depends only on the number of ions.
NaCl dissociates as Na+ and Cl- ; CsBr dissociates as Cs+ and Br-
But the concentration of the solutions are different.
Concentration (morality ) of NaCl = Moles /Litre = (1 g /58.44g/mol)/0.01L
Total number of ions in NaCl solution = 2 x (1 g /58.44g/mol)/0.01L ( 1 mol NaCl gives 2 moles ions, 1 mol Na+ and 1 mol Cl-)
= 1.71×2RT
Similarly total number of ions in CsBr solution = 2 x (1 g /212.80 g/mol)/0.01L
= 0.47×2RT
Therefore osmotic pressure is higher in NaCl solution.
A balloon has an initial volume of 2.954 L containing 5.50 moles of helium. More helium is added so that the balloon expands to 4.325 L. How much helium (moles) has been added if the temperature and pressure stay constant during this process.
Answer:
8.05 moles
Explanation:
5.50 / 2.954 = x / 4.325
x = 8.05
According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added so that the balloon expands to 4.325 L.
What is ideal gas equation?The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.
In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314 so n=V/R=4.325/8.314=0.520 moles.
Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.
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A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of the precipitate is 2.9 g. Answer the questions below to determine the percent yield. 3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq) 1. a. Which product is the precipitate? b. How many moles of the precipitate would one expect to be produced from 0.030 moles of calcium nitrate? c. How many grams of solid do you expect to be produced? d. What is the percent yield?
Answer:
a. Ca₃(PO₄)₂.
b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced
c. 3.1g of Ca₃(PO₄)₂
d. Percent yield = 93.5%
Explanation:
a. Based on the reaction:
3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)
3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.
As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,
Ca₃(PO₄)₂b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate
0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =
0.010 moles of Ca₃(PO₄)₂ can we expect to be producedc. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;
0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =
3.1g of Ca₃(PO₄)₂d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:
[tex]\frac{2.9g}{3.1g} *100[/tex]
Percent yield = 93.5%(a) The product in solid state would be the precipitate. Hence, the precipitate would be Ca3(PO4)2
(b) From the balanced equation of the reaction: 3 moles of Ca(NO3)2 is required for 1 mole of Ca3(PO4)2
If there are just 0.030 moles of Ca(NO3)2, then"
3 moles = 1
0.030 moles = 1 x 0.030/3
= 0.01 moles of Ca3(PO4)2
In other words, 0.01 moles of the precipitate would be expected to be produced from 0.030 moles of calcium nitrate.
(c) 0.01 moles solid (Ca3(PO4)2) is expected. Mass of Ca3(PO4)2 expected:
mass = mole x molar mass
molar mass of Ca3(PO4)2 = 310.18 g/mol
mass of Ca3(PO4)2 expected to be produced = 0.01 x 310.18
= 3.1018 g
Hence, 3.1018g of solid is expected to be produced.
(d) Percentage yield = actual yield/theoretical yield x 100
= 2.9/3.1018 x 100
= 93.5%
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According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrogen gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
176.8 g of ammonia, NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6 g
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34 g.
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.
This can be obtained as follow:
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.
Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.
For a spontaneous process, which of the following MUST be true?
A. TΔS>ΔH
B. ΔG>0
C. ΔSuniv>0
D. ΔSsys>ΔSsurr
Answer:
C; ΔSuniv>0
Explanation:
In this question, we want to select which of the options must be true.
What we should understand is that for a process to be spontaneous, the change in entropy must be greater than 0 i.e the change in entropy must be positive.
Looking at the options we have; option C is the correct answer.
Option B looks correct but it is wrong. This is because if change in universal entropy is greater than zero, then change in Gibbs free energy must be less than zero for spontaneity to occur
It can be deduced that for a spontaneous process, B. ΔG>0.
What is a spontaneous process?It should be noted that a spontaneous process simply means a process that occurs without input of matter or electrical energy.
In this case, for a spontaneous process, it's true that ΔG>0, it should be noted that a spontaneous process related to the second law in thermodynamics. This is characterized by an increase in entropy.
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A gas mixture contains 3.50 moles of helium, 5.00 moles of krypton and 7.60 moles of neon. A) What is the mole fraction for each gas
Answer:
.217, .311, and .472, respectively.
Explanation:
The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).
X of helium=3.50/16.10 = .217
X of krypton=5.00/16.10 = .311
X of neon=7.60/16.10 = .472
A variation of the acetamidomalonate synthesis can be used to synthesize threonine. The process involves the following steps: Ethoxide ion deprotonates diethyl acetamidomalonate, forming enolate anion 1; Enolate anion 1 makes a nucleophilic attack on acetaldehyde, forming tetrahedral intermediate 2; Protonation of the oxyanion forms alcohol 3; Acid hydrolysis yields dicarboxyamino alcohol 4; Decarboxylation leads to the final amino acid. Write out the mechanism on a separate sheet of paper, and then draw the structure of tetrahedral intermediate 2.
Answer:
See figure 1
Explanation:
For this reaction, we have the production of a carbanion as the first step. The base "ethoxide" can remove a hydrogen-producing a negative charge in the carbon (enolate anion 1). Then this negative charge can attack the carbon of the carbonyl group in the molecule acetaldehyde and the tetrahedral intermediate 2 is form. In the next step, we have the protonation of the oxygen to produce alcohol 3. A continuation we have the hydrolysis of the ester groups to produce the Dicarboxyamino alcohol and finally, we have a decarboxylation reaction we will produce the amino acid Threonine.
To further explanations see figure 1
I hope it helps!
The absorption spectrum of argon has a line at 515 nm. What is the energy of
this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's
constant is 6.626 x 10-34 Jos.)
O A. 2.59 x 1027j
O B. 3.86 x 10-28 J
O C. 3.86 x 10-19 J
O D. 2.59 x 1018 J
Answer:
OPTION C is correct
3.86 x 10-19 J
Explanation:
Energy of the line can be calculated using below formula
E= h ν.................(1)
Where E= energy
h= plank constant= 6.626 10-34 J s
c=speed of light=3 x 108 m/s
But we know that Velocity V= = c / λ
Then substitute into equation (1) we have
E = h c / λ.............(2)
We can calculate our( hc ) in nm for unit consistency
h c =( 6.626 ×10^-34)x(3×108)
h c = (1.986 x 10-16 )
hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)
E = h c / λ = (1.986 x 10-16) / 515
E = 3.86 x 10-19 J
Therefore, the Energy is 3.86 x 10-19 J
g All of the molecules below have polar bonds but only one of the molecules is a polar molecule. Which one is a polar molecule? A) C2F2 B) C2Cl4 C) CO2 D) NF3 E) CF4
Answer: [tex]NF_3[/tex]
Explanation:
Geometrical symmetry of the molecule and the polarity of the bonds determine the polarity of the molecule.
The molecule that has zero dipole moment that means it is a geometrically symmetric molecule and the molecule which has some net dipole moment means it is a geometrically asymmetric molecule.
As the molecule is symmetric, the dipole moment will be zero as dipole moments cancel each other and the molecule will be non-polar.
As the molecule is asymmetric, the dipole moment will not be zero and the molecule will be polar.
Example: [tex]NF_3[/tex]
Thus, we can say that [tex]NF_3[/tex] is a polar molecule.
There are approximately 2 × 1022 molecules and atoms in each breath we take and the concentration of CO in the air is approximately 9 ppm. How many CO molecules are in each breath we take? solution
Answer:
1.8x10¹⁷ molecules of CO are in each breath we take
Explanation:
Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.
A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.
In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:
2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =
1.8x10¹⁷ molecules of CO are in each breath we take[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take
The calculation is as follows:A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.
Now CO molecules in each breath is
[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]
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