Answer:
572. 3 g of NH3
Explanation:
Equation of the reaction: 3H2 + N2 ----> 2NH3
From the equation of reaction, 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3.
Since N2 is in excess in the given reaction, H2 is the limiting reactant.
Molar mass of H2 = 2 g/mol
Molar mass of NH3 = 17 g/mol
Therefore 3 * 2 g of H2 reacts to produce 2 * 17 g of NH3
6 g of H2 produces 34 g of NH3
101 g of H2 will produce (34 * 101)/6 g of NH3 = 572.3 g of NH3
Therefore, 572.3 g of NH3 are produced
Answer:
572.33g of NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of the H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
Molar Mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g.
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Finally, we can determine the mass of ammonia (NH3) produced by reacting 101g of H2 as follow:
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Therefore, 101g of H2 will react to produce = ( 101 x 34) / 6 = 572.33g of NH3.
Therefore, 572.33g of NH3 is produced from the reaction.
The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3 . What is the approximate thickness of the aluminum foil in millimeters?(1 ounce = 28.4g)
Answer:
18130 mm
Explanation:
Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.
First we convert the weight in ounce to grams.
If 1 ounce = 28.4g
12 ounces = 12×28.4 = 340.8 g
Next we convert the area of aluminum from ft2 to m2
1ft2= 0.0929 m2
75 ft2= 75 × 0.0929= 6.9675m2
Now density of aluminum= 2.70 gcm-3
Density= mass/volume
But volume= area× thickness
Density= mass/ area × thickness
Density × area × thickness= mass
Thickness= mass/ density × area
Thickness= 340.8g / 2.70gcm-3 × 6.9675m2
Thickness= 340.8/18.8
Thickness= 18.13 m
Since 1000 milimeters make 1 metre
Thickness= 18130 mm
A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-
Answer:
[tex]M=0.213M[/tex]
Explanation:
Hello,
In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:
[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]
[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]
[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]
We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:
[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]
Finally, we compute the molarity:
[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]
Regards.
Which best describes thermal energy? It is the difference between internal energies of two or more substances. It is the sum of internal energies of two or more substances. It is the portion of internal energy that can be transferred from one substance to another. It is the portion of potential energy that can be transferred from one substance to another.
Answer:
It is the portion of internal energy that can be transferred from one substance to another.
Thermal energy is the portion of internal energy that can be transferred from one substance to another.
What is thermal energy?Thermal energy is the energy an object posses which is as a result of particles movement within it.
It is also the internal energy system in a state of thermodynamic equilibrium which is as a result of its temperature. Thermal energy cannot be concert to useful work easily.
Therefore, thermal energy is the portion of internal energy that can be transferred from one substance to another.
Learn more about thermal energy from the link below.
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Which of the compounds below are amines?
1. H4C-NH-CH3
2. H3C-NH-C-CH3
H2C-CH3
1 +
H3C-CH2-N-CH3
CH3
N
3. H
4.
Answer:
1. H4C-NH-CH3
2. H3C-NH-C-CH3
H2C-CH3
1 +
H3C-CH2-N-CH3
CH3
N
3. H
4.
.
.
.
.
.
.
Ba(OH)2:_______.
A. 1 barium atom, 1 oxygen atom and 1 hydrogen atom.
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
C. 1 barium atom, 2 oxygen atoms and 2 hydrogen atoms.
D. 1 barium atom, 2 oxygen atoms and 1 hydrogen atom.
Answer: D
Explanation: Expand this (OH)2 you will get 2O, 2H
Hence 1Ba, 2O, 2H
Answer:
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
At 25.0 °C the Henry's Law constant for sulfur hexafluoride (SP) gas in water is 2.4x 10 M/atm Calculate the mass in grams of SFo, gas that can be dissolved in S25. ml. of water at 25.0 C and a SF, partial pressure of 1.90 atm Be sure your answer has the correct number of significant digits.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass is [tex]m = 0.0349 \ g[/tex]
Explanation:
From the question we are told that
The Henry's Law constant is [tex]k = 2.4 *10^{10} M/atm[/tex]
The volume of water is [tex]V = 525 \ ml = 0.525 \ L[/tex]
The partial pressure is [tex]P = 1.90 \ atm[/tex]
The temperature is [tex]T = 25 ^oC[/tex]
Henry's law is mathematically represented as
[tex]C = P * k[/tex]
Where C is the concentration of sulfur hexafluoride(SP)
substituting value
[tex]C = 1.90 * 2.4*10^{-4}[/tex]
[tex]C = 4.56*10^{-4} \ M[/tex]
The number of moles of SP is mathematically represented as
[tex]n = C * V[/tex]
substituting value
[tex]n = 0.525 * 4.56*10^{-4}[/tex]
[tex]n = 2.39 *10^{-4} \ moles[/tex]
The mass of SP that dissolved is
[tex]m = n * Z[/tex]
Where Z is the molar mass of SP which has a constant value of
[tex]Z = 146 g/mole[/tex]
So
[tex]m = 2.394*10^{-4} * 146[/tex]
[tex]m = 0.0349 \ g[/tex]
One of the reagents below gives predominantly 1,2 addition (direct addition) while the other gives predominantly 1,4 addition (conjugate addition). a) Which major organic product is the result of 1,2 addition? ---Select--- b) Draw the skeletal structure of major organic product A
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The correct option is reagent B
b
The skeletal structure of major organic product A is shown on the third uploaded image
Explanation:
The mechanism of the reaction for A and B are shown on the second the second reaction and looking at this we can see that the reagent that predominately gives 1,2 addition is reagent B
The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW
Answer:
Explanation:
From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer
0.25 L elution buffer = 250 mL elution butter
The breaking buffer that we use this week contains
10mM Tris = 0.01 M
150mM NaCl = 0.15 M
300mM imidazole. = 0.3 M
The stock concentration of Tris in 1M
Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;
[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 2.5 \ mL[/tex]
In NaCl, The amount of stock concentration is 5 M
so; using the same formula; we have:
[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]
[tex]V_1 = 7.5 \ mL[/tex]
From Imidazole ; the amount of stock concentration is
[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 75 \ mL[/tex]
Thus; we can have a table as shown as :
Stock concentration volume to be added Final concentration
1 M of Tris 2.5 mL 10 mM
5 M of NaCl 7.5 mL 150 mM
1 M of Imidazole 75 mL 300 mM
In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.
reasons for good care on computer
answer
1)maximise your software efficiency
2)Prevention against viruses and malware
3)Early detection of problematic issues
4)prevent data loss
5)Speed up your computer
which statements describe how chemical formulas, such as H2O, represent compounds?
Answer:
2 Hydrogen One oxygen
Explanation:
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.
In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.
In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.
This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.
Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.
Learn more about the Markovnikov rule here:
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If 196L of air at 1.0 atm is compressed to 26000ml, what is the new pressure
Answer:
7.5 atm
Explanation:
Initial pressure P1 = 1.0 ATM
Initial volume V1= 196 L
Final pressure P2= the unknown
Final volume V2= 26000ml or 26 L
From Boyle's law we have;
P1V1= P2V2
P2= P1V1/V2
P2= 1.0 × 196/26
P2 = 7.5 atm
Therefore, as the air is compressed, the pressure increases to 7.5 atm.
Fractionation of Crude Oil Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first: butane, heptadecane, dodecane, ethane, decane Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first:
butane, heptadecane, dodecane, ethane, decane
A. ethane > butane > decane > dodecane > heptadecane
B. heptadecane > > dodecane > decane butane > ethane
C. ethane > butane > decane> heptadecane >
D. dodecane butane > ethane > decane > dodecane > heptadecane
Answer:
A. ethane > butane > decane > dodecane > heptadecane
Explanation:
In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.
The product with the least weight rises to top height while the product with highest weight will move down.
For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is
ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).
Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane
Hot coffee in a mug cools over time and the mug warms up. Which describes the energy in this system? The average kinetic energy of the particles in the mug decreases. The average kinetic energy of the particles in the coffee increases. Thermal energy from the mug is transferred to the coffee. Thermal energy from the coffee is transferred to the mug.
Answer:
d
Explanation:
Answer:
D
Explanation:
Edge 2021
The osmotic pressure exerted by a solution is equal to the molarity multiplied by the absolute temperature and the gas constant . Suppose the osmotic pressure of a certain solution is measured to be at an absolute temperature o of 312. K. Write an equation that will let you calculate the molarity c of this solution.
Answer:
Explanation:
From the question, osmotic pressure exerted by a solution is equal to the MOLARITY multiplied by the absolute TEMPERATURE and the GAS CONSTANT r.
Let P = osmotic pressure,
C = molarity, then
T = absolute temperature
r=gas constant
The Osmotic pressure Equation exerted by a solution [tex]P=C*T*r[/tex]
[tex]P=CTr[/tex]
Then it was required in the question to write an equation that will let you calculate the molarity c of this solution, and this equation should contain ONLY symbols
C= molarity of the solution
P=osmotic pressure
r = gas constant
T= absolute temperature
[tex]C=P/(rT)[/tex]
The equation that will let us calculate the molarity c of this solution = [tex]C=P/(rT)[/tex]
A water tank measures 24in.×48in.×12in. Find the capacity of the water tank in cubic feet. Do not include units in your answer.
Answer: 8 (feet)
Explanation:
24 inches = 2 feet
48 inches = 4 feet
12 inches = 1 foot
To find volume you do Base * Width * Height
2*4*1 = 8
Hope this helps!
The correct answer is 8 (feet).
How to calculate ?
24 inches = 2 feet48 inches = 4 feet12 inches = 1 footTo find volume the method is Base * Width * Height
Therefore, 2*4*1 = 8
Hence, the capacity of the water tank in cubic feet is 8 feet.learn more about capacity below,
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When 1.550 gg of liquid hexane (C6H14)(C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 ∘C∘C to 38.13 ∘C∘C. Find ΔErxnΔErxn for the reaction in kJ/molkJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘CkJ/∘C.
Answer:
ΔErxn[tex]= -3.90*10^3KJ[/tex]
Explanation:
Given from the question
T1 = 25.87∘C
T2= 38.13∘C.
C= 5.73Kj/C
CHECK THE ATTACHMENT FOR DETAILED EXPLATION
What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Compare the conjugate bases of these three acids. Acid 1: hypochlorous acid , HClO Acid 2: phosphoric acid , H3PO4 Acid 3: hydrogen sulfide , HS- What is the formula for the weakest conjugate base ?
Answer:
The weakest conjugate is HClO-.
Explanation:
As a general rule, the stronger the Bronsted-Lowry acid, the weaker its conjugate base, and vice versa.
Acid 1: HClO is a strong acid, hence its conjugate base would be weak
Acid 2: H3PO4 is a weak acid, hence its conjugate base would be strong
Acid 3: hydrogen sulphide is also a moderately weak acid with a moderately strong conjugate base.
In order of increasing strengths:
HClO < H2S < H3PO4
Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water is assumed to be 1.0 g/mL that is 1.0 g/cm3. The molar mass of water is 18.0 g/mol. The atomic radius of hydrogen is 37 pm and of oxygen is 73 pm. The formula for the volume of a sphere is 4/3(r3
Answer:
The percentage of the void space out of the total volume occupied is 93.11%
Explanation:
A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.
To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom
Let's calculate this one after the other.
For the hydrogen, formula for the volume will be
[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]
where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters
Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31
we multiply this by the avogadro's number = 6.02 * 10^23
= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3
We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen
Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3
adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)
Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water
= (18g/mol)/(1 g/ml) = 18 ml/mol
Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml
Now, the percentage of void = volume of void/total volume * 100%
= 16.76/18 * 100% = 93.11%
Write a Lewis structure for each atom or ion. Draw the particle by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons and non-bonding electrons. Show the charge of the atom. Particles: S2-, Mg, Mg+2, P.
Answer:
The Lewis structure to this question can be described as follows:
Explanation:
Structure of Lewis for [tex]S^{2-}[/tex]:
The maximum number of electrons from valence in [tex]S^{2-}[/tex] is 8 (6 from S as well as 2 from negative change).
The valence electrons in the Lewis structure are placed on four sides of the atom.
Thus the structure of Lewis for [tex]S^{2-}[/tex] is as follows:
[tex]\left[\begin{array}{ccc} &. .&\\: &S&:\\&. .&\end{array}\right] ^{2-}[/tex]
Lewis Mg Structure:
Complete valence electrons are 2 in Mg.
The Lewis structure for Mg, therefore, is as follows:
[tex]\ . \\ Mg\\ \ .[/tex]
The Lewis structure for [tex]Mg^{2+}[/tex]
The maximum valence of electrons [tex]Mg^{2+}[/tex] in is= 0.
Thus, the structure for [tex]Mg^{2+}[/tex] is as follows:
[tex]Mg^{2+}[/tex]
Lewis structure for P :
The maximum number of valence electrons in P is = 5.
Thus, the structure for P is=
[tex]\ \ \ . \\ : P \ : \\[/tex]
Rubidium is ______ potassium in the periodic table. lodine is ______ bromine in the periodic table. Therefore, the rubidium ion is __________ than the potassium ion, and the iodine ion is___________ than the bromide ion. The _______ the distance between the rubidium ion and the iodide ion is the potassium ion and the bromide ion. Therefore, the energy associated with the interaction between rubidium and iodide is________ atomic radius means that than that between , and the lattice energy of potassium bromide is ________ more exothermidc.
Answer:
The given blanks can be filled with below, below, larger, larger, larger, larger and smaller.
Explanation:
In the periodic table, rubidium comes below the potassium, and iodine comes below bromine. Therefore, it can be said that the ion of rubidium is larger in comparison to potassium ion, and similarly the ion of iodine is larger in comparison to the ion of bromine.
When the atomic radius is larger it signifies that the distance in between the ion of iodine and the ion of rubidium is larger in comparison to that between the ion of potassium and the ion of bromine. Thus, smaller energy is associated with the interaction between iodine and rubidium, and potassium bromide's lattice energy is more exothermic.
Ni
Express your answer in condensed form in the order of orbital filling as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2.
Answer:
[Ar]3d^84s^2
Explanation:
From the question given, we are asked to write the condensed form of electronic configuration of nickel, Ni.
To do this, we simply write the symbol of the noble gas element before Ni in a squared bracket followed by the remaining electrons to make up the atomic number of Ni.
This is illustrated below:
The atomic number of Ni is 28.
The noble gas before Ni is Argon, Ar.
Therefore, the condensed electronic configuration of Ni is written as:
Ni(28) => [Ar]3d^84s^2
Answer:
[Ar] 4s^23d^8
Explanation:
what is the correct ionic equation, including all coefficients, charges, and phases for the following sets of reactants? Assume that the contribution of protons from H2SO4 is near 100%.
Ba(OH)2(aq)+H2SO4(aq) —>
help, I have no clue
Answer:
Ba(OH)2(aq)+H2SO4(aq) gives us 2BaH+H2O
Explanation:
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder is Pf=1.24 atm. What is the enthalpy change for this process? ΔH =
Answer:
[tex]\Delta H=-11897J[/tex]
Explanation:
Hello,
In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:
[tex]\Delta H=\Delta U+V\Delta P[/tex]
Whereas the change in the internal energy is computed by:
[tex]\Delta U=nCv\Delta T[/tex]
So we compute the initial and final temperatures for one mole of the ideal gas:
[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]
Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:
[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]
Then, the volume-pressure product in Joules:
[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]
Finally, the change in the enthalpy for the process:
[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]
Best regards.
The change in enthalpy is 70.42J
Data;
Volume of the gas = 4.86LInitial Pressure = 10.90 atmFinal Pressure = 1.24 atmChange in Enthalpy = ?Change in EnthalpyThe change of enthalpy is calculated as
[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]
The volume change is negligible
The change in enthalpy here is equal to change in internal energy over ΔE
[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]
The change in enthalpy is 70.42J
Learn more on change in enthalpy here;
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What is the conjugate acid in the following equation hbr + H2O yields h30 positive + BR negative
Answer:
HBr + H2O = H3O+ + Br-
So our conjugate acid is the H3O+ to H2O
Explanation:
A conjugate acid of a base results when the base accepts a proton.
Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.
Answer:
Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L
Explanation:
Complete Question
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
Solution
Noting that the precipitate is Copper as it is the only solid by-product of this reaction.
89 mg of Copper is produced from this reaction.
We convert this into number of moles for further stoichiometric calculations
Mass of Copper = 89 mg = 0.089 g
Molar mass of Copper = 63.546 amu
Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole
From the stoichiometric balance of the reaction,
1 mole of Copper is produced from 1 mole of Copper (II) Sulfate
0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.
Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole
Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = 0.001401 mole
Volume in L = (400/1000) = 0.4 L
Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.
Concentration in g/L = (Concentration in mol/L) × (Molar Mass)
Concentration in mol/L = 0.0035025 M
Molar mass of Copper (II) Sulfate = 159.609 g/mol
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f
Hope this Helps!!!!
The concentration of the original copper solution is 0.035 M.
The equation of the reaction is;
Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)
Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles
Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.
From the question, we are told that the volume of solution is 400.mL or 0.04L.
Hence, the concentration of the solution is; number of moles /volume
= 0.0014 moles/0.04L = 0.035 M
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Missing parts;
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
which best describes a mixture.
A it has a single composition and it has a set of characteristics
B it can have different compositions but it has a set of charactaristics that does not change
C it has a single composition but it has a set of characteristics that does change
D it can have different compositions and it has a set of characteristics that does change
Answer:
B) It can have different compositions, but it has a set of characteristics that does not change.
Explanation:
On e d g e n u i t y
I believe the answer is d lmk if im wrong or right
There are __________ moles of N atoms present in a 2.0 g C8H10O2N4.
Answer:
[tex]n_N=0.041molN[/tex]
Explanation:
Hello,
In this case, for this mole-mass relationship, we are able to compute the moles of nitrogen atoms by firstly obtaining the moles of the given compound, considering its molar mass that is 194 g/mol:
[tex]n_{C_8H_{10}O_2N_4}=2.0gC_8H_{10}O_2N_4*\frac{1molC_8H_{10}O_2N_4}{194gC_8H_{10}O_2N_4} =0.01molC_8H_{10}O_2N_4[/tex]
Then, by knowing that one mole of the given compound has four moles of nitrogen atoms, we apply the following relationship:
[tex]n_N=0.01molC_8H_{10}O_2N_4*\frac{4molN}{1molC_8H_{10}O_2N_4} \\\\n_N=0.041molN[/tex]
Best regards.
Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm
1. Calculate the volume of the bar:
2. Calculate the (experimental) density of the bar:
3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?
4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%
Answer:
1= Volume
= Length x breath x height
= 13.90 x 2.9 x 0.081
=3.26511
2= Density = Mass ÷ volume
= 11.3 ÷ 3.26511
= 3.461 (3d.p)
idk the rest because you haven't shown a picture of the rest
Answer:
1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%
Explanation:
Experimental data:
Mass = 11.3 g
Length = 13.90 cm
Width = 2.9 cm
Thickness = 0.081 cm
Calculations:
1. Volume of bar
V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³
2. Experimental density
[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]
3. Identity of metal
The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)
The metal is probably barium.
4. Percent difference
[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]