Answer:
b. radicals form easily in the presence of chlorine radicals.
Explanation:
Chlorine radicals perform the first propagation step: because "radicals form easily in the presence of chlorine radicals."
This is because the first propagation step consumes a CHLORINE RADICAL while the second propagation step regenerates a CHLORINE RADICAL. In this way, a chain reaction occurs, whereby one CHLORINE RADICAL can ultimately cause thousands of molecules of methane to be converted into chloromethane with C12 present.
Hence, in this case, the correct answer is that "radicals form easily in the presence of chlorine radicals."
Choose all the answers that apply. Silicon (Si) has 14 protons and an atomic mass of 28. Silicon has _____. three electron shells 14 electrons 14 neutrons two electron shells 28 electrons
Answer:
three electron shells
14 electrons
14 neutrons
Explanation:
Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.
Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.
The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.
Based on the "Reactivity in Substitution Reactions" experiment, which molecule would be expected to react the fastest using AgNO3 in water-ethanol ?
Answer:
C) EtOH 1% AgNO3
5. How many grams of tin metal can be produced from smelting (heating) of a 4.5 kilograms of tin (IV) oxide? (Note: Elemental tin and oxygen gas are the only products of this reaction).
Answer:
About 3500 grams of tin.
Explanation:
We want to determine amount of tin metal (in grams) that can be produced from smelting 4.5 kilograms of tin(IV) oxide.
First, write the chemical compound. Since our cation is tin(IV), it forms a 3+ charge. Oxygen has a 2- charge, so we will have two oxygen atoms. Hence, tin(IV) oxide is given by SnO₂.
By smelting it, we acquire elemental tin and oxygen gas. Hence:
[tex]\text{SnO$_2$}\rightarrow \text{Sn} + \text{O$_2$}[/tex]
(Note: oxygen is a diatomic element.)
The equation is balanced as well.
To convert from SnO₂ to only Sn, we can first convert from grams of SnO₂ to moles, use mole ratios to convert to moles of Sn, and then from there convert to grams.
Since Sn has a molar mass of 118.71 g/mol and oxygen has a molar mass of 15.999 g/mol, the molar mass of SnO₂ is:
[tex](118.71)+2(15.999) = 150.708\text{ g/mol}[/tex]
Therefore, given 4.5 kilograms of SnO₂, we can first convert this into grams using 1000 g / kg and then using the ratio:
[tex]\displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}[/tex]
We can convert this into moles.
Next, from the chemical equation, we can see that one mole of SnO₂ produces exactly one mole of Sn (and also one mole of O₂). So, our mole ratio is:
[tex]\displaystyle \frac{1\text{ mol Sn}}{1\text{ mol SnO$_2$}}[/tex]
With SnO₂ in the denominator to simplify units.
Finally, we can convert from moles Sn to grams Sn using its molar mass:
[tex]\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
With the initial value and above ratios, we acquire:
[tex]\displaystyle 4.5\text{ kg SnO$_2$}\cdot \frac{1000 \text{ g SnO$_2$}}{1\text{ kg SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol Sn}}{1 \text{ mol SnO$_2$}} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
Cancel like units:
[tex]=\displaystyle 4.5\cdot \frac{1000}{1}\cdot \displaystyle \frac{1}{150.708}\cdot \displaystyle \frac{1}{1} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1}[/tex]
Multiply. Hence:
[tex]\displaystyle = 3544.5696...\text{ g Sn}[/tex]
Since we should have two significant figures:
[tex]=3500 \text{ g Sn}[/tex]
So, about 3500 grams of tin is produced from smelting 4.5 kg of tin(IV) oxide.
Answer:
3546g
Explanation:
start w/ tin (IV) oxide n elemental tin and oxygen gas are the only products of this reaction
SnO2 -> Sn + O2
Sn molecular wt: 119
O2 molecular wt: 32
SnO2 molecular wt: 119+32 = 151
so Sn / SnO2 wt ratio = 119 / 151
4.5 kilograms of tin (IV) oxide will produce:
= 4.5 * 119 / 151
= 3.546 kg
or 3546 grams of tin metal
no need to involve moles ;)
K always has the same value
at a given temperature
regardless of the amounts of
reactants or products that
are present initially.
Select one:
True
False
hope it will help you
Why does increasing the temperature of two reactants in solution make a
reaction proceed more quickly?
Answer:
-The particles of the two reactants will gain kinetic energy and collide with one another more frequently and forcefully, which makes the reaction take place more quickly
A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
Explanation:
Ore contains ---- 65.2% [tex]TiO_2[/tex]
Mass% of titanium in TiO2 can be calculated as shown below:
[tex]mass percentage of Ti in TiO2=\frac{mass of Ti}{mass of TiO_2} *100\\=(47.86g/79.866g)* 100\\=59.9[/tex]
Given 10.0 metric tons of titanium is required.
The mass of ore that should be processed can be calculated as shown below:
Mass of Ti = ore x TiO2 % x Ti mass %
10.0 x 1000 kg = M (mass of ore) x (65.2/100) x (59.9/100) Ti
=>M=(10.0 metric tons) / (0.652 x 0.599)
=>M=25.6 metric tons
Hence, the mass of ore required is 25.6 metric tons.
The minimum amount of rutile ore to be processed is 25.6 metric tons.
Explanation:
Given :
The sample of rutile, 65.2% [tex]TiO_2[/tex] (s) by mass.
To find:
The minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
Solution:
The atomic mass of titanium = 47.867 g/mol
The atomic mass of oxygen = 15.999 g/mol
The molar mass of [tex]TiO_2[/tex] [tex]= 47.867 g/mol+2\times 15.999 g/mol=79.865 g/mol[/tex]
Percentage of the titanium in [tex]TiO_2[/tex] :
[tex]Ti(\%)=\frac{1\times 47.867 g/mol}{79.865 g/mol}\times 100\\=59.935\%[/tex]
Quantity of titanium required = 10.0 metric ton
Quantity of [tex]TiO_2[/tex] from rutile ore = x
[tex]59.935\%=\frac{10.0 \text{metric ton}}{x}\times 100\\x=\frac{10.0 \text{metric ton}}{59.935}\times 100=16.7 \text{metric ton}[/tex]
Mass of [tex]TiO_2[/tex] form rutile ore= 16.7 metric ton
The percentage of [tex]TiO_2[/tex] in rutile = 65.2 %
The quantity of rutile ore to be processed = M
[tex]65.2\%=\frac{16.7 \text{metric ton}}{M}\times 100\\M=\frac{16.7 \text{metric ton}}{65.2}\times 100=25.6 \text{metric ton}[/tex]
The minimum amount of rutile ore to be processed is 25.6 metric tons.
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A rigid, sealed container that can hold 26 L of gas is filled to a pressure of
5.97 atm at 374 °C. The pressure suddenly decreases to 3.64 atm. What is
the new temperature inside the container, in units of °C?
Answer:
121 °C
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 374 °C
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
NOTE: Volume = constant
Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 374 °C
Initial temperature (T₁) = 374 °C + 273
Initial temperature (T₁) = 647 K
Next, we shall determine the final temperature. This can be obtained as follow:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 647 K
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
P₁ / T₁ = P₂ / T₂
5.97 / 647 = 3.64 / T₂
Cross multiply
5.97 × T₂ = 647 × 3.64
5.97 × T₂ = 2355.08
Divide both side by 5.97
T₂ = 2355.08 / 5.97
T₂ = 394 K
Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
Final temperature (T₂) = 394 K
Final temperature (T₂) = 394 – 273
Final temperature (T₂) = 121 °C
Thus, the new temperature is 121 °C
You are given a metal sample that you are told is gold. Explain in a step-by-step procedure exactly how you could (a) determine if the metal is actually gold and (b) determine the purity of the gold if you know what other metals may be present. Write out your answer in a clear and well supported paragraph.
Answer:
The answer is provided below
Explanation:
To determine the metal is gold we will use the following steps
Calculate the density of the MetalTake the density of the pure goldCompare both densitiesTake a full water container
Place the metal in the container
Collect the water that spills out due to the placement of the metal
measure the mass of collected water.
Calculate the value in terms of the density of water, it will be the volume of metal.
Calculate the mass of the metal
Use the following formula to calculate the density of the metal
Density = Mass / Volume
Now compre the resulted density to the density of pure gold.
What is the electron domain geometry around N in N2CL4
Answer:
trigonal bipyramidal.
define surface are tension of liquid
The hydrological cycle refers to the circulation of water within the earth's hydrosphere in different from I. e. the liquid, solid and the gaseous forms.
A solution is made by dissolving 20 ml of acetic acid in 180ml of water. Calculate its volume concentration
Answer:
water is 9/10
chemical would be 1/10
Explanation:
180/200 would be water concentration in solution
and 20/200 would be chemical solution concentration in solution (if the chemical were to be polar and mix)
please help!
What is the definition of thermal chemistry?
a.The study of change that involves warm objects
b.The study of change that involves heat
c.The study of change that involves cool objects
d.The study of change that involves temperature
Classify each of the reactions listed below as a single-displacement, double-displacement, synthesis,
decomposition, oxidation reduction or combustion reaction.
Reaction Type
: 2Na + Cl2 → 2NaCl
: C2H4 + 3O2 → 2CO2 + 2H2O
: 2Ag2O-> 4Ag + O2
: BaCl2 + Na2SO4->BaSO4 +2NaCl
: 2AI + Fe2O3-> 2Fe + Al2O3
Suppose we have two rock samples, A and B. Rock A was subject to both physical and chemical weathering while rock B was subject to chemical weathering only. Which rock would experience more chemical weathering? Why? (2pts) (Hint: consider the effect of surface area on the rate of chemical weathering)
Answer:
Rock A will have far more chemical weathering than Rock B due to the rise in area effect
Explanation:
Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.
calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation
Answer:
The pressure is 5.62 atm.
Explanation:
An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
P= ?V= 5.005 Ln= 1.255 molR= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 273.5 KReplacing:
P* 5.005 L= 1.255 mol* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] *273.5 K
Solving:
[tex]P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}[/tex]
P= 5.62 atm
The pressure is 5.62 atm.
Why does the dehydration of an alcohol more often use concentrated sulfuric acid, H 2 S O 4 HX2SOX4, as the acid catalyst rather than dilute hydrochloric acid, H C l HCl
KAnswer:
See explanation
Explanation:
It is more common to use H2SO4 for dehydration reaction rather than HCl because HCl contains a good nucleophile,the chloride ion.
Owing to the presence of the chloride ion, a substitution reaction involving the chloride ion may also proceed also thereby affecting the elimination reaction.
Also, concentrated H2SO4 is a very good drying agent thus, as long as it is used, the alcohol substrate is completely dehydrated to yield the alkene.
Note that HCl is not a dehydrating agent.
Describe A Simple experiment that can be prepared in the laboratory to demonstrate the formation of Iron (III) Chloride from iron fillings
Answer:
Anhydrous iron(III) chloride may be prepared by treating iron with chlorine:[11]
{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}
Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.
Dissolving iron ore in hydrochloric acid
{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}
Oxidation of iron(II) chloride with chlorine
{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}
Oxidation of iron(II) chloride with oxygen
{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}
Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride. Hydrated iron(III) chloride can be converted to the anhydrous form by treatment with thionyl chloride.[12] Similarly, dehydration can be effected with trimethylsilyl chloride:[13]
{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}
Anhydrous iron(III) chloride may be prepared by treating iron with chlorine.
What is an iron filling?
Iron filings are small shavings of ferromagnetic material.
[tex]{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}[/tex]
Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.
Dissolving iron ore in hydrochloric acid.
Oxidation of iron(II) chloride with chlorine.
[tex]{\displaystyle {\ce {2FeCl_2_{(}aq){+~}Cl_2_{(}g)- > 2FeCl_3_{(}aq)}}}\\[/tex]
Oxidation of iron(II) chloride with oxygen.
Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride.
Hydrated iron(III) chloride can be converted to an anhydrous form by treatment with thionyl chloride. Similarly, dehydration can be affected by trimethylsilyl chloride.
[tex]{\displaystyle {\ce {FeCl_3.6H2O + 12 Me_3SiCl - > FeCl3 + 6 (Me_3Si)2O + 12 HCl}}}[/tex]
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how does iron I differ from iron II
Answer:
Metals tend to form positive oxidation states. Here, Iron (I) has an oxidation state of +1 while Iron (II) has an oxidation state of +2. Similarly, Lead (I) has an oxidation state of +1 while Lead(II) has an oxidation state of +2. A change in oxidation state can rather cause significant changes in the compound.
10g of a non-volatile and non-dissociating solute is dissolved in 200g of benzene.
The resulting solution boils At temperature of 81.20oC. Find the molar mass of solute.
Given that the BP of pure benzene is 80.10oC and Its elevation boiling point constant = 2.53 oC/m.
Answer: The molar mass of solute is 115 g/mol.
Explanation:
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]
OR
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Boiling point of pure solvent (benzene) = [tex]80.10^oC[/tex]
Boiling point of solution = [tex]81.20^oC[/tex]
i = Vant Hoff factor = 1 (for non-electrolytes)
[tex]K_b[/tex] = Boiling point elevation constant = [tex]2.53^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute = 10 g
[tex]M_{solute}[/tex] = Molar mass of solute = ? g/mol
[tex]w_{solvent}[/tex] = Mass of solvent = 200 g
Putting values in equation 1, we get:
[tex]81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol[/tex]
Hence, the molar mass of solute is 115 g/mol.
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
8. Using the data from question 7 what is the molar concentration of KMnO4 ?
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Answer:
7. 0.1021 M
8. 1.167 M
10. Increase in volume of water would lower the rate of reaction
Explanation:
7. What is the molar concentration of H₂C₂O₄ ?
Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.
Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V
= 1.225 × 10⁻³ mol/12 × 10⁻³ L
= 0.1021 mol/L
= 0.1021 M
8. Using the data from question 7 what is the molar concentration of KMnO₄ ?
Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.
Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V
= 14 × 10⁻³ mol/12 × 10⁻³ L
= 1.167 mol/L
= 1.167 M
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.
Somebody help me!!
Calculate the mass of 2.046L of NO2
How many molecules (or formula units) are in 138.56 g C4H10 Express your answer using four significant figures.
Answer:
dont buy cheap and off we went
If the temperature of a volume of dieal gas ncreases for 100 to 200, what happens to the average kinetic energy of the molecules?
Answer:
It increases but less than double
Explanation:
As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.
We know, the kinetic energy of an ideal gas is given by :
[tex]$V_{avg} = \sqrt{\frac{8R}{\pi M}}$[/tex]
where, R = gas constant
T = absolute temperature
M = molecular mass of the gas
From the above law, we get
[tex]$V_{avg} \propto \sqrt{T}$[/tex]
Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.
In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$[/tex]
[tex]$(V_{avg})_2 = 1.12\ (V_{avg})_1$[/tex]
Therefore, [tex]$(V_{avg})_2 > (V_{avg})_1$[/tex]
Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.
Thus, the answer is " It increases but less that double".
Gaseous ammonia chemically reacts with oxygen gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of nitrogen monoxide produced by the reaction of 1.5 moles of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Explanation:
here's the answer to your question
The absorption spectrum of neon has a line at 633 nm. What is the energy of this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's constant is 6.626 x 10-34 J·s.)
Answer:
B
Explanation:
E = hc/[tex]\lambda[/tex]. Remember, it is in meters not nanometers so you have to convert. You end up with B.
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode where Co2 (aq) is reduced to Co (s) . Assume all aqueous solutions have a concentration of 1 mol/L.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
how does lead resemble chromium?
How much BaSO4 can be formed from 196.0 g of H2SO4?
Answer:
a) You can form 466 g of BaSO₄.
Explanation:
a) Mass of BaSO4
196 g H₂SO4 × 1 mol H₂SO4
98.08 g H₂SO4
1 mol BaSO 1 mol H₂SO4 X X
466 g BaSO4
233.39 g BaSO4
1 mol BaSO4
A researcher is attempting to produce ethanol using an enzyme catalyzed batch reactor. The ethanol is produced from corn starch by first-order kinetics with a rate constant of 0.05 hr-1. Assuming the concentration of ethanol initially is 1 mg/L, what will be the concentration of ethanol (in mg/L) after 24 hours
Answer:
The correct solution is "3.32 gm/L".
Explanation:
Given:
Rate constant,
[tex]K = 0.05 \ hr^{-1}[/tex]
Time,
[tex]t = 24 \ hours[/tex]
Concentration of ethanol,
[tex]C_o= 1 \ mg/L[/tex]
Now,
The concentration of ethanol after 24 hours will be:
⇒ [tex]C_o=C\times e^{-K\times t}[/tex]
By putting the values, we get
[tex]1=C\times e^{-0.05\times 24}[/tex]
[tex]1=C\times 0.30119[/tex]
[tex]C= 3.32 \ gm/L[/tex]
Name of this product
Answer:
Explanation:
ethyl 3-methylbenzoate
