Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]
And the undergoing chemical reaction:
[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]
Next, the moles of magnesium chloride consumed by the sodium fluoride:
[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]
[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]
Thereby, the reaction quotient is:
[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic. [H3O+] = 2.6 x 10-8 M
Answer:
To calculate the [OH-] in the solution we must first find the pOH
That's
pH + pOH = 14
pOH = 14 - pH
First to find the pH we use the formula
pH = - log [H3O+]From the question
[H3O+]= 2.6 × 10^-8 M
pH = - log 2.6 × 10^-8
pH = 7.6
pH = 8
So we pOH is
pOH = 14 - 8 = 6
To find the [OH-] we use the formula
pOH = - log [OH-]6 = - log [OH-]
Find antilog of both sides
[OH-] = 1.0 × 10^-6 MThe solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7
Hope this helps you
Consider the following chemical equation: NH4NO3(s)⟶NH+4(aq)+NO−3(aq) What is the standard change in free energy in kJmol at 298.15K? The heat of formation data are as follows: ΔH∘f,NH4NO3(s)=-365.6kJmolΔH∘f,NH+4(aq)=-132.5kJmolΔH∘f,NO−3(aq)=-205.0kJmol The standard entropy data are as follows: S∘NH4NO3(s)=151.1Jmol KS∘NH+4(aq)=113.4Jmol KS∘NO−3(aq)=146.4Jmol K Your answer should include two significant figures.
Answer:
[tex]\Delta _rG=-4.3\frac{kJ}{mol}[/tex]
Explanation:
Hello,
In this case, for the given dissociation reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):
[tex]\Delta _rH=\Delta _fH_{NH^{4+}}+\Delta _fH_{NO_3^-}-\Delta _fH_{NH_4NO_3}\\\\\Delta _rH=-132.5+(-205.0)-(-365.6)=28.1kJ/mol[/tex]
Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):
[tex]\Delta _rS=S_{NH^{4+}}+S_{NO_3^-}-S_{NH_4NO_3}\\\\\Delta _rS=113.4+146.4-151.1=108.7J/mol*K[/tex]
Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the given temperature (298.15 K), we finally obtain (two significant figures):
[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=28.1kJ/mol-(298.15 K)(108.7\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\\Delta _rG=-4.3\frac{kJ}{mol}[/tex]
Best regards.
Which statements about spontaneous processes are true? Select all that apply: A spontaneous process is one that occurs very quickly.
Answer: Here are the complete options.
A spontaneous process is one that occurs very quickly. A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium. A spontaneous process is one that occurs without continuous input of energy from outside the system. A process is spontaneous if it must be continuously forced or driven.
The correct option is
A spontaneous process is one that occurs without continuous input of energy from outside the system.
A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium
Explanation:
spontaneous process is one that occurs without continuous input of energy from outside the system and occur on its own because spontaneous processes are thermodynamically favorable characterized by a decrease in the system's free energy, they do not need to be driven by an outside source of energy. Which means that the initial energy is higher than the final energy.
A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium which will result to The sign of ΔG will change from positive to negative (or vice versa) where T = ΔH/ΔS. In cases where ΔG is: negative
243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
The frequency of a signal is found to be 6389 with an uncertainty of 436 Hz. To the correct number of significant digits, it should be reported as:
Answer:
i hope it work
Explanation:
as
accurate reading range = reading ± uncertainty
so you have to say about accurate reading that its,lies in range
=(6389-436) →(6389+436)
=5953→6825
and the correct number of significant would be 3
Consider the following practical aspects of titration.
(a) how can you tell when nearing the end point in titration?
(b) What volume of NaOH is required to permanently change the indicator at the end point?
(c) If KHP sample #1 requires 19.90 mL of NaOH solution to reach an end point, what volume is required for samples #2 and #3?
(d) if vinegar sample #1 requires 29.05 mL of NaOH solution to reach an endpoint, what volume is required for samole #2 and #3?
Answer:
A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration
B) The volume of NaOH required to permanently change the indicator at the end point is a drop of NaOH
c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same
D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same
Explanation:
A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration
B) The volume of NaOH required to permanently change the indicator at the end point is a drop of NaOH
c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same
D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same
0.22 L of HNO3 is titrated to equivalence using 0.18 L of 0.2 MNaOH. What is the concentration of the HNO3?
Answer:
0.16 M
Explanation:
Data provided as per the question is below:-
Volume of [tex]HNO_3[/tex] = 0.22 L
The Volume of NaOH = 0.18 L
Morality of NaOH = 0.2
According to the given situation, the calculation of the concentration of the [tex]HNO_3[/tex] is shown below:-
For equivalence,
Number of the equivalent of [tex]HNO_3[/tex] = Number of equivalents of NaOH
[tex]= \frac{0.18\times0.2}{0.22}[/tex]
[tex]= \frac{0.036}{0.22}[/tex]
= 0.16363 M
or
= 0.16 M
How many moles of NaOH is needed to neutralize 45.0 ml of 0.30M H2SeO4? Question 2 options: A) 0.00675 B) 27.0 C) 0.027 D) 0.0135
Answer:
C) 0.027
Explanation:
In this case we can start with the reaction between [tex]NaOH[/tex] and [tex]H_2SeO_4[/tex], so:
[tex]H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O[/tex]
We have an acid ([tex]H_2SeO_4[/tex]) and a base ([tex]NaOH[/tex]), therefore we will have an acid-base reaction in which a salt is produced ([tex]Na_2SeO_4[/tex]) and water ([tex]H_2O[/tex]).
Now we can balance the reaction:
[tex]H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O[/tex]
If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can calculate the moles using the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.3~M~=~\frac{mol}{0.045~L}[/tex]
[tex]mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4[/tex]
In the balanced reaction, we have a 2:1 molar ratio between the acid and the base (for each mol of [tex]H_2SeO_4[/tex] 2 moles of [tex]NaOH[/tex] are consumed), with this in mind we can calculate the moles of NaOH:
[tex]0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH[/tex]
I hope it helps!
g Solution of barium hydroxide reacts with phosphoric acid to produce barium phosphate precipitate and water. How many mL of 6.50 M calcium hydroxide solution are required to react with a phosphoric acid solution of 45.00 mL that has a concentration of 8.70 M protons (hydrogen ions)
Answer:
30.12 mL.
Explanation:
We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:
Phosphoric acid H3PO4 will dissociate in water as follow:
H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)
From the balanced equation above,
1 mole of H3PO4 produces 3 moles of H+.
Therefore, XM H3PO4 will produce 8.70 M H+ i.e
XM H3PO4 = 8.70/3
XM H3PO4 = 2.9 M.
Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.
Next, we shall write the balanced equation for the reaction. This is illustrated below:
2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O
From the balanced equation above, we obtained the following:
Mole ratio of the acid, H3PO4 (nA) = 2
Mole ratio of the base, Ba(OH)2 (nB) = 3
Data obtained from the question include the following:
Molarity of base, Ba(OH)2 (Mb) = 6.50 M
Volume of base, Ba(OH)2 (Vb) =.?
Molarity of acid, H3PO4 (Ma) = 2.9 M
Volume of acid, H3PO4 (Va) = 45 mL
The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:
MaVa /MbVb = nA/nB
2.9 x 45 / 6.5 x Vb = 2/3
Cross multiply
2 x 6.5 x Vb = 2.9 x 45 x 3
Divide both side by 2 x 6.5
Vb = (2.9 x 45 x 3) /(2 x 6.5)
Vb = 30.12 mL
Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL
An element has an atomic number of 36, what element is it? Question 4 options: Kr K Se Es
Answer:
[tex]\Huge \boxed{\mathrm{Kr}}[/tex]
Explanation:
Krypton is an element in the periodic table with an atomic number of 36.
The symbol for Krypton is Kr.
Answer:
KR.
Explanation:
Use the periodic table for reference:
Calculate a pressure at a point that is 100 m below the surface of sea water of density 1150 kgm?
Answer:
1229.08975 kPa
Explanation:
Given that:
The depth of the water = 100 m below the surface of the water.
The density of the water = 1150 kgm
The mass of the water = depth of the water × density of the water
The mass of the water = 100 m × 1150 kgm
The mass of the water = 115000 kgf/m²
The mass of the water is the pressure of the water at a depth of 100 m below the surface of the water.
Since 1 kgf/m² = 0.00980665 kPa
Then 115000 kgm² = (115000 × 0.00980665) kPa
= 1127.76475 kPa
At standard temperature and pressure , the atmospheric pressure = 101.325 kPa.
Therefore, the pressure below the surface of sea water = 1127.76475 kPa + 101.325 kPa
= 1229.08975 kPa
Aqueous potassium nitrate (KNO3) and solid silver bromide are formed by the reaction of aqueous potassium bromide and aqueous silver nitrate (AgNO3). Write a balanced chemical equation for this reaction
Answer:
For the mentioned reaction, the balanced chemical equation is:
KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)
The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.
From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.
When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which of the following is correct for this titration?
A. Initially the pH will be less than 1.00.
B. The pH at the equivalence point will be 7.00.
C. It will require 12.50 mL of NaOH to reach the equivalence point.
When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which of the following is correct for this titration?
A. Initially the pH will be less than 1.00.
B. The pH at the equivalence point will be 7.00.
C. It will require 12.50 mL of NaOH to reach the equivalence point.
a)A, C
b) A, B
c) B, C
d) B
Answer:
c) B, C
Explanation:
NaOH(aq) + HBr(aq) -----> NaBr(aq) +H2O(l)
1) concentration of acid CA= 0.05 M
Concentration of base CB= 0.1 M
Volume of acid VA= 25.00ml
Volume of base VB= unknown
Number of moles of acid NA= 1
Number of moles of base NB= 1
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
VB= CAVANB/CB NB
VB= 0.05 × 25 × 1/ 0.1 ×1
VB= 12.5 ML
2.
Which of the examples is potassium?
es )
A)
B)
B
C)
Answer:
examples of things which contain potassium are:
green vegetables
root vegetables
fruits
potassium chloride
potassium sulphate
Explanation:
if you need a specific answer please send the options
Answer:
C
Explanation:
The answer is the one with 20 protons, 20 neutrons, and 6-8-8-2 electrons.
Compounds A and B (both C10H14) show prominent peaks in their mass spectrum at m/z 134 and 119. Compound B also shows a less prominent peak at m/z 91. On vigorous oxidation with chromic acid, compound A is nonreactive while compound B yielded terephthalic acid.
Required:
From this information, deduce the structures of both compounds, and then draw the structure of B.
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!
why are(±)-glucose and (-)-glucose both classified as D sugar
Answer:
See explanation
Explanation:
We must remember that the D / L nomenclature refers to the orientation of the hydroxyl group on carbon 5. If the "OH" is on the right we will have a D configuration. Yes, the "OH" is on the left we will have an L configuration. (See figure 1)
Now, the orientation of this "OH" has nothing to do with the ability of the molecule to deflect polarized light. If the molecule deflects light to the left we will have the symbol "(-)" (levorotation) if the molecule deflects light to the right we will have the symbol "(+)" (dextrorotation).
So in the "D" configuration, we can have both a right (+) and a left (-) deviation.
I hope it helps!
What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Learn more about an acid here: https://brainly.com/question/4519963
Enter the balanced chemical equation for the reaction of each of the following carboxylic acids with KOH.Part Aacetic acidExpress your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part B2-methylbutanoic acid (CH3CH2CH(CH3)COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part C4-chlorobenzoic acid (ClC6H4COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).
Answer:
Explanation:
Answer in attached file .
The literature value for the Ksp of Ca(OH)2 at 25 °C is 4.68E−6. Imagine you ran the experiment and got a calculated value for Ksp which was too high. Select all of the possible circumstances which would cause this result.
A. The HCl was more concentrated than the labeled molarity (0.0500 M).
B. The Ca[OH]2 solution may have been supersaturated.
C. The HCl was less concentrated than the labeled molarity (0.0500 M).
D. The Ca[OH]2 solution may have been unsaturated.
E. The titration flask may have not been clean and had a residue of a basic solution.
F. The titration flask may have not been clean and had a residue of an acidic solution.
Answer:
D. The Ca[OH]2 solution may have been unsaturated
Explanation:
The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.
If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.
Which of the following represents a compound made of five molecules? CO 5 C 2O 5 C 5O 5CO 2
Answer:
Co5
Please mark me brainliest so that I get encouraged to make more great answers like this one!
Answer:
GUYS ITS 5CO 2
Explanation:
The total kinetic energy of a body is known as:
A. Thermal energy
B. Convection
C. Potential energy
D. Temperature
The total kinetic energy of a body is known as Thermal energy. Option A
What is thermal energy?Thermal energy is the direct sum of all the available random kinetic energies of molecules.
Also note that thermal energy is directly proportional to temperature in Kelvin.
Thus, the total kinetic energy of a body is known as Thermal energy. Option A
Learn more about thermal energy here:
https://brainly.com/question/19666326
#SPJ1
Answer:
A.) Thermal energy
Explanation:
I got it correct on founders edtell
2NO + 2H2 ⟶N2 + 2H2O What would the rate law be if the mechanism for this reaction were: 2NO + H2 ⟶N2 + H2O2 (slow) H2O2 + H2 ⟶2H2O (fast)
Answer:
rate = [NO]²[H₂]
Explanation:
2NO + H2 ⟶N2 + H2O2 (slow)
H2O2 + H2 ⟶2H2O (fast)
From the question, we are given two equations.
In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.
This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.
This means our rate law is;
rate = [NO]²[H₂]
If the theoretical yield of a reaction is 332.5 g and the percent yield for the reaction is 38 percent, what's the actual yield of product in grams? \
A. 8.74 g
B. 12616 g
C. 116.3 g
D. 126.4 g
Answer: D - 126.4g
Explanation:
% Yield = Actual Yield/Theoretical Yield
38% = Actual Yield/332.5
38/100 = Actual Yield/332.5
(.38)(332.5) = 126.35 g = 126.4 g Actual Yield
Answer:
is D. the correct answer
Explanation:
I'm not sure if it is. Please let me know if I'm mistaking.
A soft drink contains 63 g of sugar in 378 g of H2O. What is the concentration of sugar in the soft drink in mass percent
Answer:
[tex]\% m/m= 14.3\%[/tex]
Explanation:
Hello,
In this case, the by mass percent is computed as shown below:
[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]
Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:
[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]
Best regards.
Question 14 (5 points)
What's the acid ionization constant for an acid with a pH of 2.11 and an equilibrium
concentration of 0.30 M?
O A) 4.87x10-8
B) 1.99x10-6
C) 3.32x10-4
OD) 2.01x10-4
Answer:
D) 2.01 x 10⁻⁴ .
Explanation:
pH = 2.11
[ H⁺ ] = [tex]10^{-2.11}[/tex]
Let the acid be HA
It will ionise as follows .
HA ⇄ H⁺ + A⁻
in equilibrium .30 [tex]10^{-2.11}[/tex] [tex]10^{-2.11}[/tex]
Acid ionisation constant Ka = [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]
= 2 x 10⁻⁴
Answer:
D) 2.01 x 10⁻⁴ is correct!
Explanation:
I got it in class!
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4NH3(g) 5O2(g)4NO(g) 6H2O(g) Using standard thermodynamic data at 298K, calculate the free energy change when 1.81 moles of NH3(g) react at standard conditions.
Answer:
-434.14 kJ
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)
Step 2: Calculate the standard free energy change (ΔG°r) for the reaction
We will use the following expression.
ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))
ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)
ΔG°r = -959.42 kJ
Step 3: Calculate the standard free energy change for 1.81 moles of NH₃
959.42 kJ are released per 4 moles of NH₃.
[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
Question is incomplete, the complete question is as follows:
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
A. Toxicity, because it can be observed by altering the state of the substance
B. Boiling point, because it can be observed by altering the state of the substance
C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms
D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms
Answer:
B.
Explanation:
A student can examine a substance without altering the bonds within the molecules by examining its boiling point.
The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.
Hence, the correct answer is "B".
discuss four factors of learning
Answer:
plz mark as BRAINLIEST plz...
Explanation:
●Intellectual factor: The term refers to the individual mental level. ...
●Learning factors: ...
●Physical factors: ...
●Mental factors: ...
When balancing redox reactions under acidic conditions, hydrogen is balanced by adding: Select the correct answer below:
a. hydrogen gas
b. water molecules
c. hydrogen atoms
d. hydrogen ions
Answer:
water molecules
Explanation:
Redox reactions are carried out under acidic or basic conditions as the case may be.
If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.
For instance, the equation for reduction of MnO4^- under acidic condition is shown below;
MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)
