In a buffer solution made of acetic acid and sodium acetate, if a small amount of acid is added, the added acid will react with whome?
Answer:
The acid reacts with the conjugate base producing more weak acid.
Explanation:
A buffer solution is defined as the mixture of a weak acid and its conjugate base or a weak base with its conjugate acid.
The acetic buffer, CH₃COOH/CH₃COO⁻, is in equilibrium with water as follows:
CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺
When an acid HX (Source of H₃O⁺) is added to the buffer, the reaction that occurs is:
CH₃COO⁻ + HX → CH₃COOH
The acid reacts with the conjugate base producing more weak acid.In fact, this is the principle of the buffer:
An acid reacts with the conjugate base producing weak acid. And the weak acid reacts with a base producing conjugate base
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
We have a buffer formed by acetic acid and sodium acetate.
What is a buffer?A buffer is a solution used to resist abrupt changes in pH when an acid or a base is added.
How are buffers formed?They can be formed in 1 of 2 ways:
By a weak acid and its conjugate base.By a weak base and its conjugate acid.Our buffer is formed by a weak acid (acetic acid) and its conjugate base (acetate ion from sodium acetate).
When an acid (HX) is added, it is neutralized by the basic component of the buffer. The generic net ionic equation is:
H⁺ + CH₃COO⁻ ⇄ CH₃COOH
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
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What happens to the rate of dissolution as the temperature is increased in a gas solution?
A.
The rate stays the same.
B.
The rate decreases.
C.
The rate increases.
D.
There is no way to tell.
Answer:
The rate decreases
Explanation:
When we dissolve a gas in a water, the process is exothermic. This implies that heat is evolved upon dissolution of a gas in water.
Recall from Le Chateliers principle that for exothermic reactions, an increase in temperature favours the reverse reaction. The implication of these is that when the temperature of the gas is increased, less gas will dissolve in water.
Hence increase in temperature decreases the rate of solubility of a gas in water.
Answer:
B.
The rate decreases.
Explanation:
Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus. The reactant is H2NCNHCH2CH2CH2CHCO minus, with an NH group, with a lone pair at the N atom, double-bonded to the first (from left to right) carbon, an NH2 group attached to the fifth carbon, an O atom double-bonded to the sixth carbon and a lone pair of electrons at the first and the second N atoms of the chain. The product has the same structure as the reactant, except that not an NH group with a lone pair, but an NH2 plus group is double-bonded to the first carbon. In addition, an NH3 plus group is attached to the fifth carbon instead of the NH2 group.
Answer:
Due to the resonance structures
Explanation:
In the question:
"Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus"
We have to take into account the structure of the amino acid arginine. In which, we have the amino and the carboxylic groups in the right and the guanidine group in the left.
In this group, we have a central carbon with three nitrogen atoms around and a double bond with the nitrogen on the top. This nitrogen on the top will accept the proton because the structure produced will have a positive charge on this nitrogen. Then, the double bond with the carbon can be delocalized into the nitrogen producing a positive charge in the carbon.
In this structure (the carbocation), we can have several resonance structures. In the blue option, we can produce a double bond with the nitrogen on the right. In the purple option, we can produce a double bond with the nitrogen on the left.
In conclusion, if the nitrogen in the top on the guanidine group accepts an hydrogen atom and we will have several resonance structures that can stabilize the molecule. Due to this, the nitrogen in the top its the best option to accept hydrogens.
See figure 1
I hope it helps!
Using only sodium carbonate, Na2CO3, sodium bicarbonate, NaHCO3, and distilled water determine how you could prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3. The total molarity of the ions should be 0.20 M. The Ka of the hydrogen carbonate ion, HCO3 - , is 4.7 x 10-11 .
Answer:
Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
Explanation:
The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:
[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:
[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex] (1)
Also, we know that:
[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex] (2)
From equation (2) we have:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex] (3)
By entering (3) into (1):
[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]
[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]
[tex] [NaHCO_{3}] = 0.103 M [/tex]
Hence, the [Na_{2}CO_{3}] is:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]
Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:
[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]
[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]
Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
I hope it helps you!
If the heat released during condensation goes only to warming the iron block, what is the final temperature (in ∘C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol and a heat capacity for iron of 0.449 J⋅g−1⋅∘C−1.)
Answer:
[tex]91°C[/tex]
Explanation:
CHECK THE COMPLETE QUESTION BELOW;
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)
Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree
From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.
Then number of moles =given mass/ molecular mass
Molecular mass of water= 18 g/mol
Given mass= 0.95 g
( 0.95 g/18 g/mol)
= 0.053 moles
Then Heat evolved during condensation = moles of water x Latent heat of vaporization
Q= heat absorbed or released
H=enthalpy of vaporization for water
n= number of moles
Q=nΔH
Q = 0.053 moles x 44.0 kJ/mol
= 2.322 Kj
=2322J
We can now calculate Heat gained by Iron block
Q = mCΔT
m = mass of substance
c = specific heat capacity
=change in temperature
m = 75 g
c = 0.450 J/g/°C
If we substitute into the above formula we have
Q= 75 x 0.450 x ΔT
2322 = 75 x 0.450 x ΔT
ΔT = 68.8°C
Since we know the difference in temperature, we can calculate the final temperature
ΔT = T2 - T1
T1= Initial temperature = 22°C
T2= final temperature
ΔT= change in temperature
T2 = T1+ ΔT
= 68.8 + 22
= 90.8 °C
=91°C
Therefore, final temperature is [tex]91°C[/tex]
The final temperature of the iron block is 91∘C.
Given that;
Heat lost during condensation of the water = Heat gained by iron block
Mass of water(mw) = 0.95 g
Latent heat of vaporization = Latent heat of condensation(L) = 44.0 kJ/mol
Mass of iron(mi) = 75.0 g
Initial temperature of iron(T1) = 22∘C
Final temperature of iron(T2) = ?
Heat capacity of iron(ci) = 0.449 J⋅g−1⋅∘C−1
So;
mwL = mici(T2 - T1)
Substituting values;
(0.95g/18g/mol) × 44.0 × 10^3(J/mol) = 75.0(g) × 0.449 J⋅g−1⋅∘C−1 (T2 - 22∘C)
2322.2 = 33.7T2 - 741.4
2322.2 + 741.4 = 37.4T2
T2 = (2322.2 + 741.4)/ 33.7
T2 =91∘C
Missing parts;
Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)
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Compound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when compound A is treated with a peroxy acid (RCO3H) followed by aqueous acid (H3O+).
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
What do we know about the unknown alkene?
We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
What is the product with the peroxyacid?
This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)
A 10.00-mL aliquot of vinegar requires 16.95 mL of the 0.4874 M standardized NaOH solution to reach the end point of the titration. Demonstrate how to calculate the molarity of the vinegar solution (HC2H3O2). Show complete work below. Answer: 0.8261 M.
Answer:
0.8261 M.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
HC2H3O2 + NaOH —> NaC2H3O2 + H2O
From the balanced equation above, we obtained the following:
Mole ratio of the acid, HC2H3O2 (nA) = 1
Mole ratio of the base, NaOH (nB) = 1
Data obtained from the question include the following:
Volume of acid, HC2H3O2 (Va) = 10 mL
Molarity of acid, HC2H3O2 (Ma) =..?
Volume of base, NaOH (Vb) = 16.95 mL Molarity of base, NaOH (Mb) = 0.4874 M
Finally, we shall determine the molarity of the acid solution, as follow:
MaVa/MbVb = nA/nB
Ma x 10 / 0.4874 x 16.95 = 1
Cross multiply
Ma x 10 = 0.4874 x 16.95
Divide both side by 10
Ma = (0.4874 x 16.95) /10
Ma = 0.8261 M.
Therefore, the molarity of the vinegar solution (HC2H3O2) is 0.8261 M.
When scientists are ready to publish the result of their experiments why is it important for them to include a description of the procedure they used
Answer: So other scientist can replicate the experiment and see if they get the same results in other words, test reliability.
Explanation:
The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?
Answer:
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Explanation:
A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.
To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:
H: 1 g/moleO: 16 g/molethe molar mass of water is:
H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole
So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?
[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]
moles of water= 0.1728
Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?
[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]
heat= 7.026 kJ
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Which of the following chemical equations corresponds to the standard molar enthalpy of formation of Na_2CO_3(s)?
a. 2 NA(s) + C(s) + 3 O(g) ------------> Na_2CO_3(s)
b. Na_2O(s) + CO_2(g) --------------->Na_2CO_3 (s)
c. Na_2(s) + C(s) + 3 O(g) -------------> Na_2CO_3 (s)
d. Na_2O(s) + CO(g) ---------------> Na_2CO_3(s)
e. 2 Na(s)+ C(s) + 3/2 O_2(g) ------------> Na_2CO_3(s)
Answer:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)
Explanation:
The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)
For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)2) What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230 mL of solution?
A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right).
Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.
You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999/gcm3. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.
Answer:
Amount of salt in 1 L seawater = 34 g
Explanation:
According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater
mass of freshwater = density * volume
1 cm³ = 1 mL
mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g
mass of freshwater + cup = 734.265 + 25 = 759.265 g
Therefore, mass of equal volume of seawater = 759.265 g
Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)
1 liter = 1000 cm³ = 1000 mL;
Density of seawater = mass / volume
Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L
Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L
mass of 1 Liter seawater = 1033.01 g
mass of 1 Liter freshwater = 999 g
mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g
Therefore, amount of salt in 1 L seawater = 34 g
g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4
Answer:
0.0002 M
Explanation:
The molarity of the HCl required would be 0.0002 M.
First, let us consider the balanced equation of the reaction:
[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]
Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.
Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:
0.0041 x 2 = 0.0082 mole HCl
Volume of the HCl = 40.95 L
Molarity = mole/volume
Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M
A small amount of solid calcium hydroxide is shaken vigorously in a test tube almost full of water until no further change occurs and most of the solid settles out. The resulting solution is:______.
Answer:
Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.
Explanation:
Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.
Small amounts of calcium hydroxide salt, [tex]Ca(OH)_{2}_(s)[/tex] is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.
I hope this explanation is helpful.
A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure (p)will be
A.2/3p
B.3/5p
C.2/5p
D.1/3p
Answer this with reason
Answer:
B.3/5p
Explanation:
For this question, we have to remember "Dalton's Law of Partial Pressures". This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.
Additionally, we have a proportional relationship between moles and pressure. In other words, more moles indicate more pressure and vice-versa.
[tex]P_i=P_t_o_t_a_l*X_i[/tex]
Where:
[tex]P_i[/tex]=Partial pressure
[tex]P_t_o_t_a_l[/tex]=Total pressure
[tex]X_i[/tex]=mole fraction
With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:
moles of hydrogen gas
The molar mass of hydrogen gas ([tex]H_2[/tex]) is 2 g/mol, so:
[tex]6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2[/tex]
moles of oxygen gas
The molar mass of oxygen gas ([tex]O_2[/tex]) is 32 g/mol, so:
[tex]64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2[/tex]
Now, total moles are:
Total moles = 2 + 3 = 5
With this value, we can write the partial pressure expression for each gas:
[tex]P_H_2=\frac{3}{5}*P_t_o_t_a_l[/tex]
[tex]P_O_2=\frac{2}{5}*P_t_o_t_a_l[/tex]
So, the answer would be 3/5P.
I hope it helps!
alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.
A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
Answer:
The pH of this solution = 5.06
Explanation:
Given that:
number of moles of CH3COOH = 0.100 mol
volume of the buffer solution = 1.0 L
number of moles of NaC2H3O2 = 0.100 mol
The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
we know that concentration in mole = Molarity/volume
Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex] = 0.10 M
The chemical equation for this reaction is :
[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]
The conjugate base is CH3COO⁻
The concentration of the conjugate base [CH3COO⁻] is = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]
= 0.10 M
where the pka (acid dissociation constant)for CH3COOH = 4.74
If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M
The ICE Table for the above reaction can be constructed as follows:
[tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]
Initial 0.10 0.035 0.10 -
Change -0.035 -0.035 + 0.035 -
Equilibrium 0.065 0 0.135 -
By using Henderson-Hasselbalch equation:
The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]
The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]
The pH of this solution = 4.74 + log (2.076923077 )
The pH of this solution = 4.74 + 0.3174
The pH of this solution = 5.0574
The pH of this solution = 5.06 to two decimal places
Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.
Answer:
There are no forces holding the gas particles together.
Explanation:
By heating a 93% pure kclo3 sample, what percentage of its mass is reduced?
2KCLO3---->2KCL+3O2
Explanation:
free your mind drink water and go outside take fresh air you will get answers
An unknown gas diffuses 5 times slower than that of H2.The moleculer mass of unknown gas is??
Answer:
50.
Explanation:
We can write Graham's Law of Diffusion as:
(Rate 1)^2 = Molecular Mass 2
-------------- -------------------------
(Rate 2)^2 Molecular Mass 1
So using the Given Information:
1^2 / (1/5)^2 = Molecular Mass of unknown gas / 2, so:
25 = M/2
M = 50.
A sample is found to contain 1.29×10-11 g of salt. Express this quantity in picograms
Answer:12.9e-12g or in short 12.9pg
Explanation:as p=1e-12
Atoms are indivisible spheres. 1.plum pudding model 2.Dalton model 3.Bohr model
Answer: 2. Dalton Model
Explanation:
John Dalton proposed that atoms are indivisible spheres. Although his model of an atom was not entirely new to the scientific world since the ancient Greeks has made a similar statement in the past ( all matter are made up of small indivisible particle called atom).
As of when Dalton proposed his model of an atom, electrons and nucleus where yet to be discovered.
1. Explain what the police siren sounds like to Jane:
2. Explain what the police siren sounds like to John:
3. Explain why the police siren sounds different between Jane and John:
Answer:
1. the siren has a lower pitch to Jane
2. the siren has a higher pitch to John
3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter
Explanation:
The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.
This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.
What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH
Answer:
0.0630
Explanation:
The molar mass of urea = 60 g/mol
we all know that:
[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
Then; the number of moles of urea
= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]
= 0.0667 mol
Similarly; the number of moles of methanol
= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]
= 0.9988 mol
The total number of moles = (0.0667 + 0.9988) mol
= 1.0655 mol
Finally,the mole fraction of urea [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]
[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]
= 0.0630
How much work (in Joules) is required to expand the volume of a pump from 0.00 L to 2.50 L against an external pressure of 1.10 atm
Answer:
W = 278.64375 Joules
Explanation:
The information given in this problem are;
Initial volume = 0L
Final volume = 2.50L
ΔV = 2.50 - 0 = 2.50 L
External pressure, P = 1.10 atm
Work = ?
These parameters are related by the equation;
w = - P ΔV
W = - (1.10 )(2.50)
W = 2.75 L atm
Upon conversion to joules;
1 liter atmosphere is equal to 101.325 joule
W = 278.64375 Joules
2. In what part of an atom can protons be found?
a. Inside the electrons
b. Inside the neutrons
C. Inside the atomic nucleus
d. Inside the electron shells
Answer:
c
Explanation:
it's found inside the atomic nucleus
Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value?
Answer:
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
Rate constant:
t(1/2) = ln 2 / K
As half-life of Cesium-137 is 30.2 years:
30.2 years = ln 2 / K
K = 0.02295 years⁻¹
Replacing this result and with the given data of the problem:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.02295 years⁻¹* t + ln[A]₀
Ln ([A] / [A₀]) = -0.02295 years⁻¹* t
As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:
Ln (0.2) = -0.02295 years⁻¹* t
70.1 years = t
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original valueAn aqueous solution of potassium bromide, KBr, contains 4.34 grams of potassium bromide and 17.4 grams of water. The percentage by mass of potassium bromide in the solution is 20 %.
Answer:
True
Explanation:
The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:
Percentage by mass = mass of substance in solution/mass of solution x 100
In this case;
mass of KBr = 4.34 grams
mass of water = 17.4 grams
mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74
Percentage by mass of KBr = 4.34/21.74 x 100
= 19.96 %
19.96 is approximately 20%.
Hence, the statement is true.
When 2 moles of NH3(g) react with N2O(g) to form N2(g) and H2O(g) according to the following equation, 880 kJ of energy are evolved. 2NH3(g) 3N2O(g)4N2(g) 3H2O(g) Is this reaction endothermic or exothermic
Answer:
Explanation:
This is a bit of a trick question.
Usually an exothermic reaction is written as
A + B - heat = C + D
The meaning of this equation is that when the bonds of the reactants break, heat has to be given away to the environment. On the left, exothermic means that heat has to be given.
The wording on this question means that heat is a product
A + B = C + D + heat.
In other words heat is given up to the environment. So this reaction is exothermic.
Complete the following equation of nuclear transmutation.
23892U + 126C → 24498Cf + 6 ______
Complete the following equation of nuclear transmutation.
U + C → Cf + 6 ______
A) 1n
B) 0 e
C) 0 e
D) 1H
E) 0g 0 -1 +1 1 0
Answer:
Option A. 1 0n
Explanation:
Details on how to balanced the equation for the reaction given in the question above can be found in the attached photo.
The missing part of the transmutation equation as it has been shown is 1/o n. Option A
What is nuclear transmutation?Nuclear transmutation is the process of shifting the number of protons in an atom's nucleus to change one element into another. Nuclear processes that change one atomic nucleus into another with a different atomic number are involved.
The production of nuclear energy, radioactive decay, and the creation of new isotopes for use in science and industry all depend on nuclear transmutation, a fundamental idea in nuclear physics.
We have the equation as;
238/92 U + 12/6 C ----> 244/98 Cf + 6 1/0 n
Learn more about nuclear transmutation:https://brainly.com/question/30078683
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Consider the compound hydrazine N2H4 (MW = 32.0 amu). It can react with I2 (MW = 253.8 amu) by the following reaction 2 I2 + N2H4 ------------- 4 HI + N2 (a) How many grams of I2 are needed to react with 36.7 g of N2H4? (b) How many grams of HI (MW = 127.9 amu) are produced from the reaction of 115.7 g of N2H4 with excess iodine?
Answer:Cobb
Explanation:What y'all
