Answer:
Mass of MgSO₄ = 37.61 g
Explanation:
Given data:
Mass of MgSO₄ needed = ?
Volume of solution = 0.250 L
Molarity of solution = 1.25 M
Solution:
Formula;
Molarity = number of moles of solute / L of solution
Now we will put the values in formula:
1.25 M = number of moles of solute / 0.250 L
Number of moles of solute = 1.25 M ×0.250 L
(M = mol/L)
Number of moles of solute = 0.3125 mol
Mass = number of moles × molar mass
Mass = 0.3125 mol × 120.366 g/mol
Mass = 37.61 g
A sailor on a trans-Pacific solo voyage notices one day that if he puts 375. mL of fresh water into a plastic cup fresh water weighing 15.0 g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup.
Required:
Calculate the amount of salt dissolved in each liter of seawater.
Answer:
Amount of salt dissolved in each liter of seawater = 40 g
Explanation:
According to Archimedes principle, a body will float in a fluid if the upthrust experienced by a body is equal to the to the weight of the body.
Also, the volume of seawater displaced equals the volume of freshwater in the cup.
From the above principle, since the freshwater and cup floats in the seawater, their combined weight equals the upthrust.
Therefore, mass of equal volume of displaced seawater = mass of freshwater + mass of cup
Mass of freshwater = density of freshwater * volume
density of freshwater = 1 g/mL; volume = 375 mL
mass of freshwater = 375 mL * 1 g/mL = 375 g
mass of seawater = 375 + 15 = 390 g
mass of salt in 375 mL seawater = mass of seawater - mass of freshwater
mass of salt = (390 - 375) g = 15 g
Since 15 g of salt are dissolved in 375 mL seawater, mass of salt in 1 L of seawater =(1000 mL/ 375) * 15g = 40 g
Therefore, amount of salt dissolved in each liter of seawater = 40 g
Someone please help me pass gen chem....
Suppose a boil water notice is sent out advising all residents in the area to boil their water before drinking it or using it for cooking. You need to boil 16.5 L of water using your natural gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 17.9% of the heat generated goes towards heating the water. Assume the density of methane is 0.668 g/L, the density of water is 1.00 g/mL, and that the water has an initial temperature of 20.4 °C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane.
Answer:
Solution-
From the question the volume of water = 18 L = 18000 mL
Now we can find the mass of water = (volume of water) * (density of water)
mass of water = (18000 mL) * (1.00 g/mL)
mass of water = 18000 g
Now we find the heat required to boil water = (mass of water) * (specific heat water) * (final temperature - initial temperature)
putting the value the heat required to boil water = (18000 g) * (4.184 J/g.oC) * (100 oC - 22.7 oC)
heat required to boil water = 5821617.6 J
heat required to boil water = 5821.62 kJ
The heat given by the combustion = (heat required to boil water) / (percent of heat taken by boiling)
Heat given by combustion = (5821.62 kJ) / (19.4 /100)
Therefore the heat given by combustion = (5821.62 kJ) / (0.194)
Heat given by combustion = 30008.35 kJ
As we know that the enthalpy of combustion of methane = 802.5 kJ/mol
The moles methane used = (Heat given by combustion) / (enthalpy of combustion of methane)
moles methane used = (30008.35 kJ) / (802.5 kJ/mol)
So the moles methane used = 37.39 mol
Now the mass methane = (moles methane used) * (molar mass methane)
The mass methane = (37.39 mol) * (16.04 g/mol)
The mass methane = 599.74 g
Now the volume methane = (mass methane) / (density of methane)
volume methane = (599.7356 g) / (0.660 g/L)
volume methane = 908.69 L
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The volume of natural gas needed to boil the water if only 17.9% of the heat is generated towards heating water is ; 918.70 L
Using the given data :
Volume of water = 16.5 L = 16500 mL
mass of water = 16500 g ( 16500 mL * 1.00 g/mL )
Density of methane = 0.668 g/L
Density of water = 1.00 g/mL
First step : determine the heat needed to boil the water
Heat required = mass of water * specific heat water * ( Δ T )
= 16500 * 4.184 * ( 100 - 20.4 ) = 5495265.6 J
= 5495.265 kJ
∴ Heat required to boil water = 5495.265 kJ
next step ; determine the heat given by combustion
heat given by combustion = ( Heat required to boil water) / ( % of heat generated )
Heat given by combustion = ( 5495.265 ) / ( 17.9 % )
= 30699.80 kJ
Enthalpy of methane combustion = 802.5 kJ/mol
∴ moles of methane used = ( 30699.80 ) / ( 802.5 ) = 38.26 mol
next ; determine the mass of methane ( natural gas )
= ( moles of methane used ) * ( molar mass )
= 38.26 * 16.04 g/mol = 613.69 g
Final step : Calculate the volume of natural gas is needed to boil the water
= mass of natural gas / density of methane
= 613.69 g / 0.668 g/L
= 918.70 L
Hence we can conclude that the volume of natural gas needed to boil water if only 17.9% of the heat is 918.70 L .
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How to separate given mixture?
Answer:
Chromatography involves solvent separation on a solid medium.
Distillation takes advantage of differences in boiling points.
Evaporation removes a liquid from a solution to leave a solid material.
Filtration separates solids of different sizes.
Explanation:
how many moles of h2 can be made from the complete reaction of 3.5 moles of al?
Given: 2Al+6HCL 2Alcl3+3h2
Answer:
From the given equation, we can see that for every 2 moles of Al, we get 3 moles of H2
So, we can say the the number of moles of H2 is 3/2 times the number of moles of Al
We are given the number of moles of Al and we have to find the number of moles of H2
We have deduced the relationship:
Moles of Al * 3 / 2 = Moles of H2
Replacing the variables with given values
3.5 * 3 / 2 = Moles of H2
Moles of H2 = 5.25 moles
5. What gases was produced from decomposing hydrogen peroxide? What non-gaseous product formed from the reaction
Answer:
H2O
Explanation:
The equation for the decomposition of hydrogen peroxide is shown below;
2H2O2(l)-----> 2H2O(l) + O2(g)
Hence, the decomposition of H2O2 yields oxygen gas and water. Water is a non gaseous product of the reaction as clearly seen in the equation above.
what are the strengths in the bonds of potassium bromide
Answer: Potassium Bromide (KBr) The Ionic bond formed between Potassium and Bromine is created through the transfer of electrons from Potassium (metal) to Bromine (nonmetal).
Explanation: this type of structure departs strongly from that expected for ionic bonding and ... whose roots go back to Max Planck's explanation in 1900 of the properties of ... types of interactions between elementary particles (the strong force, the weak force, ...
Which is the product of that reaction
Answer:
B
Explanation:
8 You are given 20.00g of a dry mixture of sand and table salt.
After adding water and filtering, you are left with wet sand on
the filter paper. The filter paper and sand is then dried and the
mass of the dry sand alone is found to be 5.00g. What was the
% sand in the original mixture?
Answer:
[tex]\%sand=25\%[/tex]
Explanation:
Hello.
In this case, given the mass of the mixture, we can define it in terms of the mass of sand and table salt as shown below:
[tex]m_{sand}+m_{salt}=20.00g[/tex]
Moreover, as after filtering, the mass of dry sand turns out 5.00 g, we can compute the % sand in the original mixture by dividing this value over the mass of the mixture as shown below:
[tex]\%sand=\frac{m_{sand}}{m_{mixture}}*100\%\\ \\\%sand=\frac{5.00g}{20.00g}* 100\%\\\\\%sand=25\%[/tex]
Best regards!
10. Predict the mass of nitrogen dioxide produced if 2.30 L of ammonia are allowed to react
with excess oxygen gas at STP?
Answer:
Mass of nitrogen dioxide produced = 4.6 g
Explanation:
Given data:
Volume of ammonia = 2.30 L
Mass of nitrogen dioxide produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Number of moles of ammonia at STP:
PV = nRT
n = PV/RT
n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K
n = 2.30 atm .L / 22.414 atm.L/mol
n = 0.1 mol
Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.
NH₃ : NO₂
4 : 4
0.1 : 0.1
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.1 mol × 46 g/mol
Mass = 4.6 g
If it takes 26.0 mL of 0.0250 M potassium dichromate to titrate 25.0 mL of a solution containing Fe2 , what is the molar concentration of Fe2
Answer:
Explanation:
moles of potassium dichromate = .0250 x .026 = 65 x 10⁻⁵ moles
1 mole of potassium dichromate reacts with 6 moles of Fe⁺²
65 x 10⁻⁵ moles of potassium dichromate will react with
6 x 65 x 10⁻⁵ moles of Fe⁺²
= 390 x 10⁻⁵ moles
390 x 10⁻⁵ moles are contained in 25 mL of solution
molarity of solution = 390 x 10⁻⁵ / 25 x 10⁻³
= 15.6 x 10⁻² M .
It took 70 seconds for 280cm³ of nitrogen to diffuse through a membrane. If Carbon(IV)Oxide is allowed to diffuse through the same membrane, how long will it take the gas to do so ?
Answer:
t = 125.3 seconds
Explanation:
Molar mass of CO2 = 12+2(16) = 66
Molar mass of N2 = 2(14)= 28
rate of diffusion of N2 = volume/ time = 280cm³/70s
= 4cm³/s
let rate of CO2 = rate of diffusion of CO2 = volume/time
= 400/t
Using Graham's law of diffusion,
rN2/rCO2 = √M(CO2)/M(N2)
4/400/t =√44/28 = 4t/400= √11/7
t/100 = 1.253 , t= (100)(1.253)
t = 125.3 seconds
hence it takes CO2 125.3 seconds to diffuse through the membrane
Label the parts of the electric circuit that best correspond to the heart, arteries, veins, and cells.
Answer:
1 ➡️ Cells
2 ➡️ Arteries
3 ➡️ Veins
4 ➡️ Heart
Explanation:
The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.
The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.
From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood
The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.
What is an electric circuit?The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.
In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.
In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.
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What does chemical equations and chemical formulas have in common?
Answer:
Chemical symbols refer to chemical elements only. They do not necessarily refer to atoms of that element, but also to ions.
Explanation:
When nitrogen, oxygen, fluorine, sodium, magnesium and aluminum ionize, they all will have:
a. different electron configuration from each other.
b. an unchanged electron configuration.
c. the same charge.
d. the same electron configuration (isoelectronic) as neon.
[Definition: The word isoelectronic means that when you write out the electron configuration they are the same. An exam would be He and Li whereby both of them have 2 electrons and therefore they are both are 1s2 in their electron configurations.]
Answer: d. the same electron configuration (isoelectronic) as neon.
Explanation:
Isoelectronic species are defined as the molecules which have the same number of electrons.
Atomic number of nitrogen is 7 and thus has 7 electrons. Nitrogen has electronic configuration of 2,5 and thus can gain 3 electrons and thus [tex]N^{3-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of oxygen is 8 and thus has 8 electrons. Oxygen has electronic configuration of 2,6 and thus can gain 2 electrons and thus [tex]O^{2-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of flourine is 9 and thus has 9 electrons. Flourine has electronic configuration of 2,7 and thus can gain 1 electron and thus [tex]F^{-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of sodium is 11 and thus has 11 electrons. Sodium has electronic configuration of 2,8,1 and thus can lose 1 electron and thus [tex]Na^{+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of magnesium is 12 and thus has 12 electrons. Magnesium has electronic configuration of 2,8,2 and thus can lose 2 electrons and thus [tex]Mg^{2+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of aluminium is 13 and thus has 13 electrons. Aluminium has electronic configuration of 2,8,3 and thus can lose 3 electrons and thus [tex]Al^{3+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
the egents erosion and explain how each of them causes erosion
Answer:
Wind, water, gravity, and ice
Explanation:
Water can erode soil material. Especially if the soil is bare, dry and erodible erosion via rain particles can occur. Wind is another factor that cause soil erosion. Dry soil particles( Especially if they are fine) can move to other areas if wind exists. Ice is another issue on erosion. Ice, when it is melting, can carry soil particles.
I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.
Answer:
175
Explanation:
Which neutral atom is isoelectronic with Cl-??
And we can see that the potassium ion, K+, has the same electronic configuration as the chloride ion, Cl-, and the same electronic configuration as an atom of argon, Ar. Therefore, Ar, Cl-, and K+ are said to be isoelectronic species.
3. A student took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and
found it read 196.5 9. What was the student's percent error?
Answer:
The answer is 1.71 %Explanation:
The percentage error of a certain measurement can be found by using the formula
[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]
From the question
actual mass = 200 g
error = 200 - 196.59 = 3.41
We have
[tex]p(\%) = \frac{3.41}{200} \times 100 \\ = 1.705[/tex]
We have the final answer as
1.71 %Hope this helps you
(ii) What is the approximate chlorine-carbon-chlorine bond angle in C2Cl4?
Answer:
∠
O
−
C
−
C
l
≅
120
∘
The central carbon is
s
p
2
−
hybridized........
Explanation:
And thus
∠
C
l
−
C
−
C
l
and
∠
O
−
C
−
C
l
≅
are
120
∘
to a first approximation.
Why this value? We look to
VSEPR theory
. There are 3 regions of electron density around the central carbon, and the most stable geometry is a trigonal plane. While there is a
carbonyl
group, i.e. a
C
=
O
bond, the
π
bond is conceived to lie above and below this trigonal plane.
The
carbonyl oxygen
is likewise conceived to be
s
p
2
-hybridized
, however, here, there are 2 lone pairs on the oxygen centre.
The approximate chlorine-carbon-chlorine bond angle in C2Cl4 is 120°.
The bond angle is defined as the angle between ant two bonds emanating from a common atom.
The compound C2Cl4 is tetrachloroethene. The carbon atoms are sp2 hybridized in this molecule.
Recall that the bond angle of an sp2 hybridized carbon atom is 120°. Therefore the chlorine-carbon-chlorine bond angle in C2Cl4 is 120°.
From the perspective of the VSEPR theory, the geometry of each carbon atom in C2Cl4 is trigonal planar which implies a bond angle of 120°.
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A. Express each Fraction in Decimal form. Round off your answers into
thousandths place.
1. 3/4
Answer:
0.75
Explanation:
3/4=0.75 so therefore its 0.75
how many moles are in a 4.2 gram gold sample
The smallest form of matter that still retains the properties of an element
Answer:
atom
Explanation:
the atom is the smallest form.
Which is a chemical property of milk
A. Milk has a ph ranging from 6.4 to 6.8
B. Milk spoils when left unrefrigerated
C. Milk boils at about 212F
D. Milk curdles when mixed with vinegar
Answer:
C. Milk boils at about 212F
Explanation:
The principal constituents of milk are water, fat, proteins, lactose (milk sugar) and minerals (salts). Milk also contains trace amounts of other substances such as pigments, enzymes, vitamins, phospholipids (substances with fatlike properties), and gases.
If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitation reaction?
2 NaOH(aq) + Cd(NO₃)₂(aq) → Cd(OH)₂ (s) + 2 NaNO₃(aq)
Answer:
[tex]m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]
Explanation:
Hello.
In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:
[tex]n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2[/tex]
Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:
[tex]m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]
Best regards.
The mass of Cd(OH)₂ that would be formed is 36.6 g
From the question,
We are to determine the mass of Cd(OH)₂ that would be formed
The given balanced chemical equation for the reaction is
2NaOH(aq) + Cd(NO₃)₂(aq) → Cd(OH)₂ (s) + 2NaNO₃(aq)
This means
2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂ and 2 moles of NaNO₃
Now, we will determine the number of moles of each reactant present
For NaOHMass = 20.0 g
Molar mass = 39.997 g/mol
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Number of moles of NaOH present = [tex]\frac{20.0}{39.997 }[/tex]
Number of moles of NaOH present = 0.50 mole
For Cd(NO₃)₂Volume = 0.750 L
Concentration = 1.00 M
From the formula
Number of moles = Concentration × Volume
∴ Number of moles of Cd(NO₃)₂ present = 1.00 × 0.750
Number of moles of Cd(NO₃)₂ present = 0.750 mole
Since,
2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂
Then,
0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂
∴ The number of moles of Cd(NO₃)₂ that reacted is 0.25 mole
Now,
From the balanced chemical reaction
2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂
Then,
0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂ to produce 0.25 mole of Cd(OH)₂
∴ Number of moles of Cd(OH)₂ formed is 0.25 mole
Now, for the mass of Cd(OH)₂ that would be formed
From the formula
Mass = Number of moles × Molar mass
Molar mass of Cd(OH)₂ = 146.43 g/mol
∴Mass of Cd(OH)₂ formed = 0.25 × 146.43
Mass of Cd(OH)₂ formed = 36.6075 g
Mass of Cd(OH)₂ formed≅ 36.6 g
Hence, the mass of Cd(OH)₂ that would be formed is 36.6 g
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How many grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution?
17.5 grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution.
What is Solution?
A solution is a homogenous mixture composed of two or more substances. In a solution, the components are evenly distributed throughout the mixture, resulting in a uniform appearance and properties.
The substance that is present in the largest amount is called the solvent, while the other substances present in lesser amounts are called solutes. When the solute dissolves in the solvent, the resulting mixture is called a solution.
To calculate the grams of NaCl needed to prepare a 35.0% salt solution, we can use the formula:
grams of NaCl = (percent salt / 100) x grams of solution
grams of NaCl = (35.0 / 100) x 50.0
grams of NaCl = 0.35 x 50.0
grams of NaCl = 17.5
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A teaspoon of salt, NaCl has a mass of about
5.0 g. How many formula units are in a
teaspoon of salt?
Answer: The answer is 5.15x10^22
Explanation:
The formula unit present in a teaspoon of salt [tex]NaCl[/tex] having a mass of about 5.0 g is [tex]5.15 \times10^{22}[/tex] formula units.
Molar mass, also known as molecular weight, is the mass of one mole of a substance. It is calculated by summing up the atomic masses of all the atoms in a molecule. The unit of molar mass is grams per mole (g/mol).
Now, to determine the number of formula units in a teaspoon of salt (NaCl), we need to use Avogadro's number and the molar mass of NaCl.
Avogadro's number [tex](N_a)[/tex] is approximately. [tex]6.022 \times10^{23}[/tex] formula units per mole.
The molar mass of [tex]NaCl[/tex] is the sum of the atomic masses of sodium (Na) and chlorine ([tex]Cl[/tex]), which are approximately 22.99 g/mol and 35.45 g/mol, respectively.
To calculate the number of formula units in 5.0 g of [tex]NaCl[/tex], we can follow these steps:
Now, calculate the number of moles of [tex]NaCl[/tex] using its molar mass:
Moles = Mass / Molar mass
Moles = [tex]5.0 g[/tex] / [tex](22.99 g/mol + 35.45 g/mol)[/tex]
Calculate the number of formula units using Avogadro's number:
Formula units = [tex]Moles \times Avogadro's number[/tex]
Let's perform the calculation:
Molar mass of [tex]NaCl[/tex]= [tex]22.99 g/mol + 35.45 g/mol = 58.44 g/mol[/tex]
Moles of [tex]NaCl[/tex] = [tex]5.0 g[/tex] / [tex]58.44 g/mol[/tex] ≈ [tex]0.0856 mol[/tex]
Formula units = [tex]0.0856 mol \times (6.022 \times 10^{23})[/tex] formula units/mol ≈ [tex]5.15 \times10^{22}[/tex]formula units.
Therefore, there are approximately [tex]5.15 \times10^{22}[/tex] formula units in a teaspoon of salt ([tex]NaCl[/tex]) having mass [tex]5.0 g[/tex].
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If 0.4743 moles of H2O are produced, how many grams of VOCl3 will also be produced?
(V2O5 + 6 HCl → 2 VOCl3 + 3 H2O)
Answer:
i hope this helps. sorry if it totally doesn't
1) The speed constant for the second order reaction in the gas phase
It varies with the temperature according to the table below. Calculate the activation energy for the process, according to Arhhenius' equation
Answer:
41.7 kJ/mol
Explanation:
ln(k) = ln(A) − Eₐ/(RT)
Pick any two points. I'll choose 100°C and 400°C.
When T = 100°C = 373 K, k = 1.10×10⁻⁹ L/mol s:
ln(1.10×10⁻⁹) = ln(A) − Eₐ/(R × 373)
When T = 400°C = 673 K, k = 4.40×10⁻⁷ L/mol s:
ln(4.40×10⁻⁷) = ln(A) − Eₐ/(R × 673)
Subtract the two equations and solve:
ln(4.40×10⁻⁷) − ln(1.10×10⁻⁹) = -Eₐ/(R × 673) + Eₐ/(R × 373)
5.991 = 0.00120 Eₐ/R
Eₐ/R = 5013.4
Eₐ = 41700 J/mol
Eₐ = 41.7 kJ/mol
If 5.00g of iron metal is reacted with 0.950g of Cl2 gas, how many grams of ferric chloride (FeCl3) will form?
Answer:
1.45g of FeCl3
Explanation:
The equation of the reaction is given as;
2Fe + 3Cl2 --> 2FeCl3
2 mol of Fe reracts with 3 mol of Cl2 to form 2 mol of FeCl3
Upon converting to mass using;
Mass = Number of moles * Molar mass
( 2 * 55.85 = 111.7g ) of Fe reacts with ( 3 * 71 = 213g ) of Cl2 to form ( 2 * 162.2 = 324.4g) of FeCl3
Cl2 is the limiting reactant as it determines how much of FeCl3 is formed
213g of Cl2 = 324.4g of FeCl3
0.950g of Cl2 = x
x = (0.950 * 324.4 ) / 213
x = 1.45g of FeCl3
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Rank the following elements in order from least to most number of moles of atoms in a 10.0 g sample: Sn, Si, Se, S
Answer:
[tex]\rm Sn[/tex], [tex]\rm Se[/tex], [tex]\rm S[/tex], [tex]\rm Si[/tex].
Explanation:
The relative atomic mass of an element is numerically equal to the mass (in grams) of one mole of its atoms. This quantity can help estimate the number of moles of atoms in each of these four [tex]10.0\; \rm g[/tex] samples.
Look up the relative atomic mass for each of these four elements (on a modern periodic table.)
[tex]\rm Si[/tex]: [tex]28.085[/tex].[tex]\rm S[/tex]: [tex]32.06[/tex].[tex]\rm Se[/tex]: [tex]78.971[/tex].[tex]\rm Sn[/tex]: [tex]118.710[/tex].The relative atomic mass of [tex]\rm Si[/tex] is (approximately) [tex]28.085[/tex]. Therefore, the each mole of silicon atoms would have a mass of approximately [tex]28.085\; \rm g[/tex]. How many moles of silicon atoms would there be in a [tex]10.0\; \rm g[/tex] sample?
Given:
[tex]m(\rm Si) = 10.0\; \rm g[/tex]. [tex]M(\mathrm{Si}) = 28.085\; \rm g \cdot mol^{-1}[/tex].Number of mole of silicon atoms in the sample: [tex]\displaystyle n(\mathrm{Si}) = \frac{m(\mathrm{Si})}{M(\mathrm{Si})} = \frac{10.0\; \rm g}{28.085\; \rm g \cdot mol^{-1}}\approx 0.356\; \rm mol[/tex].
Similarly:
[tex]\displaystyle n(\mathrm{S}) = \frac{m(\mathrm{S})}{M(\mathrm{S})} = \frac{10.0\; \rm g}{32.06\; \rm g \cdot mol^{-1}}\approx 0.312\; \rm mol[/tex].
[tex]\displaystyle n(\mathrm{Se}) = \frac{m(\mathrm{Se})}{M(\mathrm{Se})} = \frac{10.0\; \rm g}{78.971\; \rm g \cdot mol^{-1}}\approx 0.127\; \rm mol[/tex].
[tex]\displaystyle n(\mathrm{Sn}) = \frac{m(\mathrm{Sn})}{M(\mathrm{Sn})} = \frac{10.0\; \rm g}{118.710\; \rm g \cdot mol^{-1}}\approx 0.0842\; \rm mol[/tex].
Therefore, among these [tex]10.0\; \rm g[/tex] samples:
[tex]n(\mathrm{Sn}) < n(\mathrm{Se}) < n(\mathrm{S}) < n(\mathrm{Si})[/tex].
It is not a coincidence that among these four samples, the one with the fewest number of atoms corresponds to the element with the largest relative atomic mass.
Consider two elements, with molar mass [tex]M_1[/tex] and [tex]M_2[/tex] each. Assume that [tex]n_1[/tex] moles and [tex]n_2[/tex] moles of atoms of each element were selected, such that the mass of both samples is [tex]m[/tex]. That is:
[tex]m = n_1\cdot M_1[/tex].
[tex]m = n_2\cdot M_2[/tex].
Equate the right-hand side of these two equations:
[tex]n_1 \cdot M_1 = n_2\cdot M_2[/tex].
[tex]\displaystyle \frac{n_1}{n_2} = \frac{M_2}{M_1} = \frac{1/M_1}{1/M_2}[/tex].
In other words, the number of moles atoms in two equal-mass samples of two elements is inversely proportional to the molar mass of the two elements (and hence inversely proportional to the formula mass of the two elements.) That explains why in this question, the sample containing the smallest number of atoms corresponds to element with the largest relative atomic mass among those four elements.