Answer:
Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules / one mole of O₂F₂)
Explanation:
The Avogadros constant gives the the number of specified entities in one mole of a substance. One mole of any substance contains 6.02 × 10²³ particles. Therefore, one mole of O₂F₂ contains 6.02 × 10²³ molecules.
Also, the molar mass of a substance is the mass in grams of one mole of that substance. It is obtained by summing the relative atomic masses of all the atoms of the elements in the substance. For O₂F₂, the molar mass = (2 × 16 + 2 × 19) g/mol = 70 g/mol
Converting to molecules of O₂F₂;
To convert from grams of a substance to molecules of that substance, multiply by the ratio of one mole and mass of one mole, and then by the number of molecules per mole.
Molecules of A = mass of A × (1 mole / mass of one mole) × (6.02 × 10²³ molecules / 1 mole)
Therefore,Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules /one mole of O₂F₂)
Why does the temperature stop rising while ice melts into water?
A. The temperature does not stop rising.
B. The electrons are increasing in energy levels.
C. Because no more heat is being added to the system.
D. The energy is being absorbed to separate the particles.
Answer:
When you heat ice, its temperature rises, but as soon as the ice starts to melt, the temperature stays constant until all the ice has melted. This happens because all the heat energy goes into breaking the bonds of the ice's crystal lattice structure.
Explanation:
The temperature stop rising while ice melts into water because, the energy is being absorbed to separate the particles. This is because of latent heat of fusion.
What is latent heat of fusion?The amount of energy needed to convert the solid substance into a liquid substance by modifying its physical effects. It exists also named enthalpy of fusion. When heat exists supplied to ice, it begins melting and heat is used to increase temperature initially. But after the occasional temperature of ice does not vary and the extra heat exists utilized to melt the ice by cracking bonds between crystal lattice of ice.
The temperature stops increasing while the ice melts into the water because the energy exists being absorbed to divide the particles. This exists because of the latent heat of fusion.
Therefore, (D) option is the correct answer.
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What is the main reason for using a data table to collect data?
A. To interpret the possible meaning of the data
B. To find the possible errors that were made in recording the data
C. To organize the information so that it is easier to understand
O
D. To make an experimental journal more attractive
Answer:
c
Explanation:
table of data help us to understand and present our work better
I did it and got it right, it's c
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
Given the following balanced equation:
3Cu(s) + 8HNO3(aq) = 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
Determine the mass of copper (II) nitrate that would be formed from the complete reaction
of 35.5g of copper with an excess of nitric acid.
Answer: The mass of copper (II) nitrate produced is 105.04 g.
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)
Given mass of copper = 35.5 g
Molar mass of copper = 63.5 g/mol
Plugging values in equation 1:
[tex]\text{Moles of copper}=\frac{35.5g}{63.5g/mol}=0.560 mol[/tex]
The given chemical equation follows:
[tex]3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)[/tex]
By the stoichiometry of the reaction:
If 3 moles of copper produces 3 moles of copper (II) nitrate
So, 0.560 moles of copper will produce = [tex]\frac{3}{3}\times 0.560=0.560mol[/tex] of copper (II) nitrate
Molar mass of copper (II) nitrate = 187.56 g/mol
Plugging values in equation 1:
[tex]\text{Mass of copper (II) nitrate}=(0.560mol\times 187.56g/mol)=105.04g[/tex]
Hence, the mass of copper (II) nitrate produced is 105.04 g.
When determining the amount of oxidant present by titration, you can use iodine/starch as an indicator. First, the oxidant, like hypochlorite, oxidizes Choose... When starch and iodine are both present, the solution is Choose... During the titration, a titrant like thiosulfate reduces the
The question is incomplete, the complete question is;
When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.
First, the oxidant, like hypochlorite, oxidizes
Choose...
neutral iodine into iodide ion
iodide ion into neutral iodine
iodate polyatomic ion into iodide ion
When starch and iodine are both present, the solution is
Choose...
blue-black
brownish yellow
clear
During the titration, the titrant, like thiosulfate, reduces the
Choose...
iodide ion into iodate polyatomic ion
neutral iodine into iodide ion
iodide ion into neutral iodine
When the iodine has completely reacted at the endpoint of the titration, the solution should become
Choose...
clear
blue-black
brownish yellow
Answer:
1. iodide ion into neutral iodine
2. blue-black
3. neutral iodine into iodide ion
4. clear
Explanation:
Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;
ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)
When iodine is added, the colour of the starch solution immediately changes to blue-black.
A reduction reaction occurs when the titrant, thiosulfate is added as follows;
I2 + 2S2O32- → 2I- + S4O62-
The solution at end point is found to become clear again.
When should a line graph be used
Answer:
Line graphs are used to track changes over short and long periods of time. When smaller changes exist, line graphs are better to use than bar graphs. Line graphs can also be used to compare changes over the same period of time for more than one group.
Use the reaction: 2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2HNO3(aq) What volume (mL) of 0.568 M AgNO3(aq) is needed to form 0.21 g of Ag2SO4(s)
Answer:
The mole ratio of AgNO3 to Ag2SO4 IS 2:1 .0.657 g Ag2SO4 x 1 mol / 312 g = 0.00211 mol Ag2SO4.
0.00211 mol Ag2SO4 x 2 mol AgNO3 / 1 mol Ag2SO4 = 0.00421 mol AgNO3
0.00421 mol AgNO3 x 1 L / 0.123 mol AgNO3 = 0.0342 L = 34.2 mL of AgNO3 solution.Therefore,34.2ml of 0.123M AgNO3 will be required.
What separates the inner planets from the outer planets?
a. Main asteroid belt
b. Main comet belt
c. Kuiper belt
d. Outer planet belt
please help this is for SCIENCE test I need help
Answer:
main asteroid belt separates the inner planets from the outer planets
An unknown compound has the following chemical formula:
Co(OH),
where x stands for a whole number.
Measurements also show that a certain sample of the unknown compound contains 5.1 mol of oxygen and 2.59 mol of cobalt.
Write the complete chemical formula for the unknown compound.
since we are given the moles for Co and O, we'll divide both of those moles by the lowest mole quantity, which is, in this case, 2.59. After dividing, we see that the ratio of O to Co is 2:1. So, for every 1 Co atom, there has to be 2 O atoms. we can then insert the 2 in for OH to satisfy this ratio.
LION
If 3.0L of helium at 20°C is allowed to expand to 4.4L, with pressure remain the same
Answer:
This question is asking to find the new temperature
The answer for the final temperature is 429.73K
Explanation:
Using Charles law equation as follows:
V1/T1 = V2/T2
Where;
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question;
V1 = 3.0L
V2 = 4.4L
T1 = 20°C = 20 +273 = 293K
T2 = ?
Using V1/T1 = V2/T2
3/293 = 4.4/T2
Cross multiply
293 × 4.4 = 3 × T2
1289.2 = 3T2
T2 = 1289.2 ÷ 3
T2 = 429.73K
What is the molecule shown below?
A. Pentane
B. Trimethylethane
C. 2,2-dimethylpropane
D. 3-dipropane
Q2
Answer:
C
Explanation:
if we were to followw the IUPAC
How long do spent fuel rods remain dangerously radioactive?
Answers
A.
The rods are no longer radioactive because the radioisotopes are used up.
B.
Spent fuel rods remain radioactive for several years after the fuel is exhausted.
C.
It takes tens of thousands of years for the radioisotopes in the rods to decay to safe levels.
D.
It is impossible to determine how long it will take for the radioisotopes to decay because they last too long.
Answer:
c
Explanation:
it takes 10,000 years to just reduce down the decay
consider the following thermochemical reaction for kerosene
2C12H26+37O2=24CO2+15026kj.
a. when 21.3g of CO2 are made, how much heat is released?
b. if 500.00kj of heat are released by thye reaction, how many grams of C12H26 have been consumed.?
c. if this reactionwere being used to generate heat, how many grams of C12H26 would have to be reacted to generate enough heat to raise the temperature of 750g of liquid water from 10 degrees celcius to 90 degrees celcius
Thermochemistry has to do with heat evolved or absorbed in a chemical reactions. Thermochemical equations are equations in which the heat of reaction is included in the reaction equation. The reaction of moles and heat of reaction is important here.
This question has to do with thermochemistry and thermochemical equations.
The answers to each of the questions are shown below;
a) 300.52 KJ
b) 11.39 g
c) 5.78 g
The equation of the thermochemical reaction is;
2C12H26 + 37O2-------> 24CO2 + 15026KJ
Number of moles of CO2 released = 21.3g/44g/mol = 0.48 moles
From the reaction equation;
15026KJ is released when 24 moles of CO2 is released
x KJ is released when 0.48 moles of CO2 is released
x = 15026KJ * 0.48 moles/24 moles
x = 300.52 KJ
b) If 2 moles of C12H26 released 15026KJ of heat
x moles of C12H26 released 500.00KJ
x = 2 * 500.00KJ/15026KJ
x = 0.067 moles
Mass of C12H26 consumed = 0.067 moles * 170 g/mol = 11.39 g
c) Heat gained by water = heat released by combustion of kerosene
Heat gained by water = 0.75 Kg * 4200 * (90 -10)
Heat gained by water = 252 KJ
If 2 moles of C12H26 produced 15026KJ
x moles of C12H26 produces 252 KJ
x = 2 * 252/15026
x = 0.034 moles
Mass of C12H26 = 0.034 moles * 170 g/mol = 5.78 g
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A chemical equation is shown below.
FeC13 + NaOH -> NaCl + Fe(OH)3
What are the coefficients that should be added to balance this question?
Explain how this chemical reaction demonstrates the conservation of mass.
Answer:
Explanation:
Balance the OHs first. There's 3 on the right.
FeC13 + 3NaOH -> NaCl + Fe(OH)3
Now balance the Nas. There are 3 on the left.
FeC13 + 3NaOH -> 3NaCl + Fe(OH)3
The equation is now balanced
Chemical Left Right
Fe 1 1
Na 3 3
Cl 3 3
OH 3 3
In order to complete the reaction of hexyl magnesium bromide with acetone, what next step needs to be done.
a. Fractional Distillation.
b. Vacuum filtration.
c. Aqueous workup.
d. Crystallization.
Answer:
Aqueous workup.
Explanation:
The reaction of hexyl magnesium bromide with acetone yields a tertiary alcohol. There is an organic phase and an aqueous phase.
Aqueous workup is the process of recovering the pure tertiary alcohol from the organic phase of the system.
Hence, in order to complete the reaction of hexyl magnesium bromide with acetone, aqueous workup is required.
The metal sample suspected of being aluminum is warmed and then submerged into water, which is near room temperature. The final temperature of the water and the metal is given below. The specific heat capacity of water is 4.18 J/g.oC. Calculate the specific heat capacity of the metal based on the data below. Remember heat lost = heat gained.
Type of metal used:
Trial 1 Trial 2 Trial 3
Mass of metal, g 2.746 g 2.750 g 2.900 g
Mass of water, g 15.200 g 15.206 g 15.201 g
Initial Temp. of Water, oC 24.7 oC 24.6 oC 24.5 oC
Initial Temp. of Metal, oC 72.1 oC 72.2 oC 71.9 oC
Final Temp of Water & Metal,oC 26.3 oC 26.2 oC 24.7 oC
ΔT for water, oC ______ ______ ______
ΔT for metal, oC ______ ______ ______
Specific heat capacity of metal, J/g.oC ______ ______ ______
Average specific heat capacity, J/g .oC ______ (use two significant figures due to ΔT of water)
Answer:
Average specific heat capacity of metal = 0.57 J/g°C
Explanation:
Heat lost = Heat gained
Heat energy gained or lost, H = mcΔT
where m = mass of substance, c = specific heat capacity, ΔT = temperature change
Trial 1:
Heat lost by metal = -[2.746 g × c × ΔT]
ΔT = (26.3 - 72.1) °C = -45.8 °C
Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C
Heat gained by water = 15.200 × 4.18 × ΔT
ΔT = (26.3 - 24.7) = 1.6 °C
Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J
From Heat lost = Heat gained
c × (125.7688)g°C = 101.6576 J
c = 101.6576 J / 125.7688 g°C
c = 0.8083 J/g°C
Trial 2:
Heat lost by metal = -[2.750 g × c × ΔT]
ΔT = (26.2 - 72.2)°C] = - 46 °C
Heat lost by metal = -[2.750 g × c × (-46 °C)
Heat lost by metal = c × (126.5) g°C
Heat gained by water = 15.206 × 4.18 × ΔT
ΔT = (26.2 - 24.6) = 1.6 °C
Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J
From Heat lost = Heat gained
c × (126.5)g°C = 101.6977 J
c = 101.697728 J / 126.5 g°C
c = 0.8039 J/g°C
Trial 3:
Heat lost by metal = -[2.900 g × c × ΔT]
ΔT = (24.7 - 71.9)°C] = - 47.2 °C
Heat lost by metal = -[2.900 g × c × (- 47.2 °C)
Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C
Heat gained by water = 15.201 × 4.18 × ΔT
ΔT = (24.7 - 24.5) = 0.2 °C
Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J
From Heat lost = Heat gained
c × (136.88)g°C = 12.708036 J
c = 12.708036 J / 136.88 g°C
c = 0.0928 J/g°C
Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3
Average specific heat capacity of metal = 0.57 J/g°C
A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at .Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to significant digits.
Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:
[tex]M=58g[/tex]
Explanation:
From the question we are told that:
Heat Capacity [tex]H=0.897[/tex]
Mass of water [tex]M=200g[/tex]
Initial Temperature of Aluminium [tex]T_a=85.6[/tex]
Initial Temperature of Water [tex]T_{w1}=16.0[/tex]
Final Temperature of Water [tex]T_{w2}=16.0[/tex]
Generally
Heat loss=Heat Gain
Therefore
[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]
[tex]M=58g[/tex]
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
A tree is an example
of a vascular plant that
is
because it
has deep roots.
A. tall
B. tiny
C. small
Dyshort
explain hydrogen dioxide
Answer:
Two molecules of hydrogen combine with two molecules of oxygen to form hydrogen peroxide. Hence, its chemical formula is H2O2. It is the simplest peroxide (since it is a compound with an oxygen-oxygen single bond). Hydrogen peroxide has basic uses as an oxidizer, bleaching agent and antiseptic
Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
Which species is the conjugate base of H2SO3
Explanation:
As you know, the conjugate base of an acid is determined by looking at the compound that's left behind after the acid donates one of its acidic hydrogen atoms.
The compound to which the acid donates a proton acts as a base. The conjugate base of the acid will be the compound that reforms the acid by accepting a proton.
In this case, sulfurous acid has two protons to donate. However, the conjugate base of sulfurous acid will be the compound left behind after the first hydrogen ion is donated.
how many moles of KF are present in 46.5 grams of KF
Explanation:
here's the answer to your question
Answer:
0.8017
Explanation:
Find the molar Mass of KF
K = 39
F = 19
Total = 58
Note: these numbers are approximate. Use your periodic table to get the exact numbers.
mols = given mass / molar mass
given mass = 46.5
molar mass = 58
mols = 46.5 / 58
mols = 0.8017
How many moles of p are needed to react with 30.1 moles of O2 SHOW the math below.
Answer:
information is missing
Explanation
reaction is needed to solve the problem
Calculating the expected pH of the buffer solution: Given that the pKa for Acetic Acid is 4.77, calculate the expected pH of the buffer solutions using the Henderson-Hasselbalch equation and the concentrations of Acetic Acid and Acetate added to the 250 ml Erlenmeyer flask: pH
Answer:
[tex]pH=4.77[/tex]
Explanation:
From the question we are told that:
pKa for Acetic Acid [tex]pK_a= 4.77[/tex]
Therefore
For Equal Concentration of acetic acid and acetatic ion
[tex]CH_3COOH=CH_3COO^-[/tex]
Generally the Henderson's equation for pH value is mathematically given by
[tex]pH=pK_a+log\frac{base}{acid}[/tex]
[tex]pH=4.77+log\frac{CH_3COO^-}{CH_3COOH}[/tex]
[tex]pH=4.77+log1[/tex]
[tex]pH=4.77[/tex]
Briefly workout the relationship between these constants:
[tex]{ \bf{K _{sp} \: and \: K _{c} }}[/tex]
In consideration of the decopmposition of hydrogen iodide.
[tex]{ \sf{2HI _{(g)} →H _{2(g)} +I _{2(g)} }}[/tex]
[tex]{ \tt{any \: help \: is \: appreciated}}[/tex]
Kc require (aqueous/gaseous) products to be on the numerator and (aqueous/gaseous) reactants to be in the denominator, whereas Ksp will require (aqueous) products to be on the numerator and (aqueous) reactants to be in the denominator. Both require products on top and reactants in the bottom.
K = [products] / [reactants]
Kc is used when a reaction reaches dynamic equilibrium, whereas Ksp is used when an insoluble ionic solid dissolved by a tiny amount in a solution, as well as in determining whether or not a precipitate will form.
Kc can be used to measure equilibrium concentration for all reactions, whereas Ksp is limited to only ionic compounds' solubility.
The decomposition of HI (g) will required the use of Kc since the species are all gaseous, and gases cannot be ionic.
Which redox reaction would most likely occur if zinc and copper metal were
added to a solution that contained zinc and copper ions?
Click for a reduction potential chart
A. Cu + Zn → Cu2+ + Zn2+
B. Cu + Zn2+
Cu2+ + Zn
C. Cu2+ + Zn → Cu + Zn2+
D. Cu2+ + Zn2+ → Cu + Zn
Answer:
C
Explanation:
b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex] is the redox reaction. Hence, option C is correct.
What is Redox Reaction?A chemical reaction taking place between an oxidizing substance and a reducing substance.
The oxidizing substance is used to lose electrons in the reaction, and the reducing substance is used to gain electrons.
On the reduction potential chart, zinc is a stronger oxidizing agent than, Copper (Cu), which is a reducing agent as compared to silver
The redox reaction most likely occurs if silver and copper metal were added to a solution that contained silver and copper ions is ;
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex]
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A student was given a solid containing a mixture of nitrate salts. The sample completely dissolved in water, and upon addition of dilute HCl , no precipitate formed. The pH was lowered to about 1 and H2S was bubbled through the solution. No precipitate formed. The pH was adjusted to 8 and H2S was again bubbled in. This time, a precipitate formed. Which compounds might have been present in the unknown?
a. Ca(NO3)2
b. AgNO3
c. Fe(NO3)3
d. Cr(NO3)3
e. Cu(NO3)2
f. KNO3
g. Bi(NO3)2
Answer:
Fe(NO3)3, Cr(NO3)3, Co(NO3)3
Explanation:
According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.
Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.
The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.
The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.
How many grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride?
Answer:
it is 11.55 and ik because I just had that question
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
Let's consider the following balanced equation.
4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)
The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:
[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]
The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:
[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
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You need to make an aqueous solution of 0.121 M magnesium acetate for an experiment in lab, using a 250 mL volumetric flask. How much solid magnesium acetate should you add
Answer:
4.27 g
Explanation:
Number of moles = concentration × volume
Concentration = 0.121 M
Volume = 250 mL
Number of moles = 0.121 M × 250/1000 L
Number of moles = 0.03 moles
Number of moles = mass/molar mass
Mass= Number of moles × molar mass
Mass= 0.03 moles × 142.394 g/mol
Mass = 4.27 g
