Answer:
See detailed explanation.
Explanation:
Hello there!
In this case, according to the given description, it turns out possible for us to infer that the second fractionating column on top of the first one will favor the light product, in this case hexane as it has the lowest boiling point and molar mass; in such a way, we can tell the following:
a) The separation between hexane and heptane will be increased as a purer hexane-rich product will be obtained on the top of the second column.
b) Will be increased as well, because the second column will remove more heptane.
c) Also, more pure heptane will be obtained on the bottom of the two columns, yet the most favored yield will be that of hexane.
All of the aforementioned is possible due to the fact that the second column will remove the amount of heptane that could not be removed on the top of the first column by taking the vapor-liquid equilibrium further from the first column's maximum separation, which is known as distillation sequences.
Regards!
compared to an atom of C-14, an atom of C-12 has a lesser
atomic number
number of protons
number of electrons
number of neutrons
Answer:
mass number
Explanation:
because the mass
number is the number of protons plus the number of neutron and the number of proton in an elements is always the same , therefore and atom of C-14 has greater mass numberhow can we convert plastic garbage energy into electric energy
Answer:
Unfortunately, we don`t know how to convert plastic material into electricity yet. I suppose an idea is for someone to invent a machine similar to biomass, where dead plants created energy, but here it`s plastic. The only issue is that it could release deadly chemicals.
Sorry if this isn`t much help, but there isn`t really an answer. :/
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.
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
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.
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
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water?
Answer:
1087.84 J
Explanation:
From the question given above, the following data were obtained:
Mass of metal (Mₘ) = 70 g
Temperature of metal (Tₘ) = 80 °C
Mass of water (Mᵥᵥ) = 100 g
Temperature of water (Tᵥᵥ) = 22 °C
Equilibrium temperature (Tₑ) = 24.6 °C
Heat lost by metal (Qₘ) =?
NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Thus, we shall determine the heat gained by water. This can be obtained as follow:
Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
Qᵥᵥ = 100 × 4.184 (24.6 – 22)
Qᵥᵥ = 418.4 × 2.6
Qᵥᵥ = 1087.84 J
Thus, the heat gained by water is 1087.84 J.
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Qᵥᵥ = 1087.84 J
Qₘ = 1087.84 J
Therefore, the heat lost by the metal is 1087.84 J
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
What is a calorimeter?A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.
Let's use the following expression to calculate the heat absorbed by the water.
Qw = c × m × ΔT
Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ
where,
Qw is the heat absorbed by the water.c is the specific heat capacity of water.m is the mass of water.ΔT is the change in the temperature for water.According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.
Qw + Qm = 0
Qm = -Qw = -10.8 kJ
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
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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.
The turbines in a hydroelectric plant are fed by water falling from a 50 m height. Assuming 91% efficiency for conversion of potential to electrical endrgy, and 8% loss of the resulting power in transmission, what is the mass flow rate of water required to power a 200 W light bulb?
From the information given;
the height of the water stream = 50 mthe efficiency of conversion from potential energy to electrical energy is 91%loss of power transmission = 8%To determine the mass flow rate, let's start by understanding some concepts and parameters.
The power is known to be the energy per unit of time. Mathematically, it can be written as:
[tex]\mathbf{Power = \dfrac{Energy}{Time}}[/tex]
[tex]\mathbf{P =\dfrac{E_p}{time}}[/tex]
[tex]\mathbf{P =\dfrac{m\times g\times z}{time}}[/tex]
where;
[tex]\mathbf{E_p}[/tex] is the potential energy of the streamm = mass flow rateg = acceleration under gravityz = heightThus;
[tex]\mathbf{E_p}[/tex] = m × 9.81 m/s² × 50 m
[tex]\mathbf{E_p}[/tex] = m × 490.5 (m²/s²)
Recall that:
The power P = 200 W, and;the conversion of the P.E = 91% = 0.91∴
[tex]\mathbf{E_p}[/tex] = 0.91 × 490.5m (m²/s²)
[tex]\mathbf{E_p}[/tex] = 446.355m (m²/s²)
Since the resulting power transmission is said to be 8%
Then;
the loss in the power transmission (P) = 100% - 8% × 446.355m (m²/s²)
the loss in the power transmission (P) = 92% × 446.355m (m²/s²)
the loss in the power transmission (P) = 0.92 × 446.355m (m²/s²)
the loss in the power transmission (P) = 410.65m (m²/s²)
Finally;
P = 410.65m (m²/s²)
[tex]\mathbf{P = 410.65 \times m (\dfrac{m^2}{s^2})}[/tex]
replacing the values, we have:
[tex]\mathbf{200 = 410.65 \times m (\dfrac{m^2}{s^2})}[/tex]
[tex]\mathbf{m = \dfrac{200 watt}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]
[tex]\mathbf{m = \dfrac{200 \dfrac{J}{s}}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]
since 1 J/s = 1 kgm²/s²)
Then:
[tex]\mathbf{m = \dfrac{200 \dfrac{\dfrac{kg\times m^2}{s^2}}{s}}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]
[tex]\mathbf{m = \dfrac{200 \ {kg}}{410.65 \ s}}[/tex]
mass flow rate of the water (m) = 0.487 kg/s
Therefore, we can conclude that the mass flow rate of the water required to power a 200 W bulb light is 0.487 kg/s
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19. Which type of chemical process is used to remove salt from ocean water?
O A. Alkylation
O B. Doping
O C. Dehydrogenation
D. Desalination
Answer:
D
Explanation:
Desalination
Removing salt from sea water is known as desalination
molecular weight of K2SO3
Explanation:
the molecular weight of K2SO3 is 158. 2598 m/s.
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.
Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?
Answer:
The percent abundance of oxygen-18 is 1.9066%.
Explanation:
The average atomic mass of oxygen is given by:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
Where:
m: is the atomic mass
%: is the percent abundance
Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:
[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]
[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]
[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]
Hence, the percent abundance of O-18 is:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]
[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]
Hence, the percent abundance of oxygen-18 is:
[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]
Therefore, the percent abundance of oxygen-18 is 1.9066%.
I hope it helps you!
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.
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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Carbon dioxide gas is collected at 27.0 oC in an evacuated flask with a measured volume of 30.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.480atm. Calculate the mass and number of moles of carbon dioxide gas that were collected.
Answer:
[tex]M_{CO_2}= 25.7g[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=27.0[/tex]
Volume [tex]V=30L[/tex]
Pressure [tex]P=0.480atm[/tex]
Generally the equation for Ideal gas is mathematically given by
PV=nRT
Therefore
[tex]n=\frac{0.480 x 30}{0.08205 x 300}[/tex]
[tex]n=0.59moles[/tex]
Generally Mass of CO2 is given as
[tex]M_{CO_2}= 0.59 * 44 g/mol[/tex]
[tex]M_{CO_2}= 25.7g[/tex]
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.
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
There are three isotopes of carbon. They have mass number of 12, 13 and 14. The average atomic mass of carbon is 12.0107 amu. What does this say about the relative abundances of the three isotopes?
Answer:
lots more of the carbon 12 than the others
havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14
Explanation:
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]
Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ
How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Answer:
endet nach selam nw
4gh7
Write a balanced half-reaction for the oxidation of manganese ion (mn2 ) to solid manganese dioxide (mno2) in acidic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s) + 4H+(aq) + 2e-
Explanation:
Step 1: Data given
The oxidation number of manganese ion (Mn2+ ) is +2
The oxidation number of manganese dioxide is +(MnO2)4
This means the oxidation number from Mn will go from +2 to +4, since it's increased, this is an oxidation reaction
Mn2+(aq) ⇒ MnO2(s)
We have to balance both sides. Mn is already the same. But on the right side we have O atoms. T obalance both sides we have to add O atoms to the left side. This by adding 2x H2O
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s)
Now the amount of O atoms is balanced, but we have H- atoms at the left side. To balance we have to add 4 H atoms to the right side
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s) + 4H+(aq)
Now the amount of atoms is balanced at both sides. We also have to check if the charge on both sides is the same.
Since the left side has a charge of +2, and right has a charge of +4, we have to add 2 electrons to balance this.
Mn2+(aq) + 2H2O(l) ⇒ MnO2(s) + 4H+(aq) + 2e-
A mixture of argon and neon gases at a total pressure of 874 mm Hg contains argon at a partial pressure of 662 mm Hg. If the gas
mixture contains 12.0 grams of argon, how many grams of neon are present?
Answer:
6.684g
Explanation:
Here, we can use the mole ratio of the gases to calculate.
We know that the mole ratio of the gases equate to their number of moles.
Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol
Thus, the number of moles produced is 5.98/32 = 0.186875
Where do we move from here?
We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.
449/851 = 0.186875/n
n =(0.186875 * 851)/449
n = 0.3542
Now we do the same for argon to get the number of moles of argon.
Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.
P = 851 - 449 = 402 mmHg
We now use the mole ratio relation.
402/851 = n/0.3542
n = (402 * 0.3542) / 851
n = 0.1673
Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.
The atomic mass of argon is 39.948 amu
The mass is thus 39.948 * 0.1673 = 6.684g
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
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
Consider the molecule PF5.
Indicate how many lone pairs you would find on the central atom:
Indicate how many total bonds are connected to the central atom (count single bonds as 1 bond, double bonds as 2 bonds, and triple bonds as 3 bonds):
Explanation:
here's the answer to your question
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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Determine whether the statement about identifying a halide is true: Regardless of any concentration of ammonium solution, the precipitates in the reaction solution of my unknown halide after 0.1M AgNO3 remain because my unknown halide solution contains Br. Select one: True False
Answer:
False
Explanation:
The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE
This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺
Activation energy is:
A. The energy needed to begin breaking the bonds of reactants.
B. None of these.
C. The maximum amount of energy reactants can hold.
D. The energy needed to begin breaking the bonds of products.
Activation energy is the energy needed to begin breaking the bonds of reactants. Hence, option A is correct.
What is activation energy?Activation energy is defined as the minimum amount of energy necessary to initiate a chemical reaction.
Hence, activation energy is the energy needed to begin breaking the bonds of reactants.
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15. In the image given below, magnesium metal is coiled as a thin ribbon. What property of metal is exhibited by it? A Ductility B Lustrous C Sonorous D Malleability
Answer: The property of magnesium that is exhibited by it is DUCTILITY. The correct option is A.
Explanation:
Magnesium is a member of the alkaline earth metals. It occurs in nature, only in the combined state, as Epsom salt, dolomite and in many trioxosilicates( IV) including talc and asbestos. They have the following physical properties:
--> Appearance: they are silvery-white solids
--> Relative density: It has a relative density of 1.74
--> DUCTILITY: it's very ductile in nature
--> melting point: it has a melting point of 660°C.
--> Conductivity: They are good conductor of heat and electricity.
Furthermore, DUCTILITY is the physical property of a metal associated with the ability to be hammered thin or stretched into wire without breaking. A metal such as magnesium can therefore be coiled as a thin ribbon without fracturing due to its ductile physical properties.
