Answer:
(a) [tex]Ksp=4.50x10^{-7}[/tex]
(b) [tex]Ksp=1.55x10^{-6}[/tex]
(c) [tex]Ksp=2.27x10^{-12}[/tex]
(d) [tex]Ksp=1.05x10^{-22}[/tex]
Explanation:
Hello,
In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:
(a) [tex]BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}[/tex]
In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}[/tex]
(B) [tex]Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}[/tex]
In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}[/tex]
(C) [tex]NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}[/tex]
In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:
[tex]Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}[/tex]
(D) [tex]La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}[/tex]
In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:
[tex]Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}[/tex]
Best regards.
A glass cylinder contains 2 gases at a pressure of 106 kPa. If one gas is at 7 kPa, what is the pressure of attributed to the other gas? a) 9 kPa b) 99 kPa c) 113 kPa d) 7 kPa e) 2 kPa (URGENT)
Answer:
b) 99 kPa
Explanation:
According to Daltons law of partial pressure, the total pressure of a mixture of two or more non reactive gases is the sum of their individual pressures. Let the total pressure of a mixture of n number of gases be [tex]P_{total}[/tex] and their individual pressure be [tex]P_1,P_2,P_3,\ .\ .\ .\ ,\ P_n[/tex], According to Daltons partial pressure law:
[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n[/tex]
Since A glass cylinder contains 2 gases at a pressure of 106 kPa, therefore n = 2. Also one gas ([tex]P_1[/tex]) is at 7 kPa. Using Daltons partial pressure law:
[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n\\P_{total}=P_1+P_2\\106\ kPa=7\ kPa+P_2\\P_2=106\ kPa-7\ kPa\\P_2=99\ kPa[/tex]
A student mixes wants to prepare 24.1 mmol of benzamide from benzoyl chloride and NH4OH. If the student uses excess 15 M NH4OH, how many mL of Benzoyl chloride must be used
Answer:
2.81mL
Explanation:
Based on the reaction:
C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl
Benzoyl chloride + ammonia → Benzamide
1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.
Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.
As molar mass of benzoyl chloride is 141g/mol, mg you require are:
mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.
to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:
3.3981g × (1mL / 1.21g) =
2.81mLGive the electron configuration for the following atoms using appropriate noble gas inner core abbreviation: Bi Cr Sr P 2. Give a set of 4 possible quantum numbers for the most energetic electron(s) of: Bi Cr Sr P n = l = ml = ms = 3. What is the symbol and name of the element with the following electron configuration?
Answer:
See explanation
Explanation:
The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;
Bismuth;
[Xe]4f14 5d10 6s2 6p3
Chromium;
[Ar]4s1 3d5
Strontium;
[Kr]5s2
Phosphorus;
[Ne]3s2 3p3
2.
Bi
6p- n=6, l= 1, ml= 1, ms= 1/2
Cr
3d- n=3, l=2, ml=2,ms=1/2
Sr
5s- n=5, l=0, ml=0, ms=1/2
P
3p- n=3, l= 1, ml= 1, ms=1/2
3.
a) Tin (Sn) - [Kr] 5s2 4d10 5p2
b) Caesium (Cs)- [Xe] 6s1
c) Copper (Cu)- [Ar] 4s1 3d10
Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm
please.
Answer:
3.00 L
Explanation:
Convert the pressure to Pascals.
P = 82 atm × (101325 Pa/atm)
P = 8,308,650 Pa
Convert temperature to Kelvins.
T = 27°C + 273
T = 300 K
Use ideal gas law:
PV = nRT
(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)
V = 0.00300 m³
If desired, convert to liters.
V = (0.00300 m³) (1000 L/m³)
V = 3.00 L
Answer:
[tex]\large \boxed{\text{3.0 L}}[/tex]
Explanation:
[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]
A reaction has the rate law, rate = k[A]3[B]. Which change will cause the greatest increase in the reaction rate?
Answer:
Increasing the concentration of A will cause the greatest change over the rate.
Explanation:
Hello,
In this case, considering the given rate law, which is third-order with respect to A, changing its concentration, the rate will be significantly modified. For instance, suppose a concentration of A and B of 1M and a symbolic rate constant (k), this causes the rate to be:
[tex]r=k[1M]^3[1M]=1k\frac{M^4}{s}[/tex]
Then, if we change the concentration of A to 2 M holding the concentration of B in 1 M, the new rate constant will be:
[tex]r=k[2M]^3[1M]=8k\frac{M^4}{s}[/tex]
Nevertheless, if we hold the concentration of A in 1 M and the concentration of B is now 2 M (same change), the new rate constant is:
[tex]r=k[1M]^3[2M]=2k\frac{M^4}{s}[/tex]
It means that increasing the concentration of A will cause the greatest change over the rate.
Best regards.
The change in concentration of A will cause the greatest increase in the reaction rate.
Rate of reaction :Given reaction has rate law ,
[tex]rate=k[A]^{3} [B][/tex]
which is third-order with respect to concentration of A
From rate law equation, It is observed that the degree of concentration of A is three.
It means that small change in concentration of A result large change in rate of reaction.Thus, the change in concentration of A will cause the greatest increase in the reaction rate.Learn more about the rate of reaction here :
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which of the following compounds exhibits dipole -dipole forces as its strongest attraction between molecules? o2,ch3Br,CCl4,He,BrCH2CH2OH
Answer:
B. CH3Br
Explanation:
Dipole -Dipole interactions take place in polar molecules.
CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.
While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.
Hence, the correct option is B. CH3Br.
The compound that exhibits dipole -dipole forces is CH3Br
Dipole -Dipole interactions:It should be taken place in polar molecules. Also, CH3Br should be the strongest attraction that lies between the molecules since it is treated as the polar molecule because of the slightly negative. While on the other hand, O2 should be non-polar molecule because of the linear structure.
Therefore, The compound that exhibits dipole -dipole forces is CH3Br
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If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×10−13.) Express your answer to two significant figures and include the appropriate units. nothing nothing
Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
Identify the elements that have the following abbreviated electron configurations.
A) [Ne] 3s23p5.
B) [Ar] 4s23d7.
C) [Xe] 6s1.
Answer:
A) Chlorine (Cl)
B) Cobalt (Co)
C) Caesium (Cs)
Hope this helps.
The abbreviated electron configurations that was given in the question belongs to
Chlorine (Cl)
Cobalt (Co)
Caesium (Cs) respectively.
Electronic configurations can be regarded as the electronic structure, which is the way an electrons is arranged in energy levels towards an atomic nucleus.The electron configurations is very useful when describing the orbitals of an atom in its ground state.To calculate an electron configuration, we can put the periodic table into sections, and this section will represent the atomic orbitals which is the regions that house the electrons. Groups one of the period table and two belongs to s-block, group 3 through 12 belongs to the d-block, while 13 to 18 can be attributed to p-block ,The rows that is found at bottom are the f-blockTherefore, electron configurations explain orbitals of an atom when it is in it's ground state.
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A calibration curve constructed from absorbance values of solutions containing a known concentration of permanganate ions has the following best-fit line:
y = (3.62× 10^3 L/mol) x
where y is the absorbance of the solution at 525 nm and x is the concentration of MnO4- (aq) in mol/L. The path length of the cuvettes used in the experiment is 1 cm. Based on this information, what is the molar absorptivity of MnO4- (aq) at 525 nm?
Answer:
3.62×10³ L/mol
Explanation:
Beer-Lambert law relates the absorbance of a sample and its concentration. Its formula is:
A = ε×C×l
Where A is absorbance of the sample, ε is molar absorptivity (A constant f each sample), C its concentration and l is path length
Now, the formula obtained was:
y = (3.62×10³ L/mol) x
Where Y ia absorbance = A, x its concentration = C and 1cm is path length.
You can write:
A = (3.62×10³ L/mol)×C×l
That means, molar absorptivity of your sample under the meaured conditions is:
3.62×10³ L/mol
Calculate the molality of a solution prepared by dissolved 19.9 g of kcl in 750ml of water
First, we find the molar mass of KCl which is AK+ACl=39+35.5=74.5g/mole
now we find out the number of moles by dividing the given mass to the molar mass n=m/M=19.9/74.5=0.26 moles. The molarity of a solution is equal to the number of moles divided by the volume of the solution. =0.26moles/0.75liters=0.346M
Which one of the following reactions must be carried out in an electrolytic cell rather than in an electrochemical cell?
a. Zn2+ + Ca → Zn + Ca2+
b. Al3+ + 3Br– → Al + (3/2)Br2
c. 2Al + 3Fe2+ → 2Al3+ + 3Fe
d. H2 + I2(s) → 2H+ + 2I–
Answer:
b. Al3+ + 3Br– → Al + (3/2)Br2
Explanation:
If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.
However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.
b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]
Electrolytic Cell v/s Electrochemical Cell:
Electrochemical cells convert chemical energy into electrical energy or vice versa while an electrolytic cell is a type of electrochemical cell in which electrical energy is converted into chemical energy.Electrochemical cells consist of a cathode (+) and an anode(-) while Electrolytic cells consist of a positively charged anode and a negatively charged cathode.Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex] is carried out in an electrolytic cell rather than in an electrochemical cell.
As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells
one is a reduction half-cell, the other is an oxidation half-cell.In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.
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Which is one characteristic of producers?
They recycle nutrients.
They do not eat other organisms.
They break down waste for energy.
They use other organisms for energy.
Answer: They use other organisms for energy
Explanation:
Answer:
they use other organisms for energy
Explanation: d
The NMR spectrum of your final compound will contain extra peaks that were not present in your starting material. For what hydrogen nuclei do those peaks occur?
Answer:
The peaks are registered from tetramethyl silane (TMS)
Explanation:
Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.
Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.
I hope this helped.
A student obtained a clean flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 34.232 g. She then filled the flask with water and found the new mass to be 60.167 g. The temperature of the water was measured to be
Answer:
25.99mL is the volume internal volume of the flask
Explanation:
To complete the question:
The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask
The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.
To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:
Mass water = Mass filled flask - Mass of clean flask
Mass water = 60.167g - 34.232g
Mass water = 25.935g of water.
To convert this mass to volume:
25.935g × (1mL / 0.997992g) =
25.99mL is the volume internal volume of the flaskWhat are the solutions to the quadratic equation 2x2 + 10x - 48 = 0?
Answer:
x = 3 , x= -8Explanation:
[tex]2x^2+10x-48\\=2\left(x^2+5x-24\right)\\x^2+5x-24\\=\left(x^2-3x\right)+\left(8x-24\right)\\=x\left(x-3\right)+8\left(x-3\right)\\=\left(x-3\right)\left(x+8\right)\\=2\left(x-3\right)\left(x+8\right)\\2\left(x-3\right)\left(x+8\right)=0\\x-3=0\\x = 0+3\\x = 3\\x+8=0\\x+8-8=0-8\\x=-8\\x=3,\:x=-8[/tex]
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was reacted with excess oxygen and 6.8g of product was obtained. What was the percent yield of the reaction?
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
4 moles of Al produce 2 moles of Al₂O₃
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
8.7g of Al₂O₃ can be produced (Theoretical yield)Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
78.2%Convert the following measurement
Answer:
6.9 Kg/mol•dL
Explanation:
To convert 6.9×10⁴ g/mol•L to kg/mol•dL,
First, we shall convert to kg/mol•L.
This can be achieved by doing the following:
Recall: 1 g = 1×10¯³ Kg
1 g/mol•L = 1×10¯³ Kg/mol•L.
Therefore,
6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³
6.9×10⁴ g/mol•L = 69 Kg/mol•L
Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.
This is illustrated below:
Recall: 1 L = 10 dL
1 Kg/mol•L = 1×10¯¹ Kg/mol•dL
Therefore,
69 Kg/mol•L = 69 × 1×10¯¹
69 Kg/mol•L = 6.9 Kg/mol•dL
Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.
Which Carbon is the triple bound attached to in 6-ethyl-2-octyne?
-first
-fourth
-third
-second
Answer:
-second
Explanation:
6-ethyl-2-octyne is an unsaturated compound with a triple bond.
6-ethyl-2-octyne will have a triple bound attached to the second carbon. The suffix -yne suggests that compound carry a triple bond and the number "2" before suffix refers to the position of triple bond that is second carbon.
Hence, the correct option is "-second ".
What is the pH of a 0.02M solution of sodium acetate (pka=4.74) to which you add HCl to a final concentration of 0.015M?
Answer:
pH = 5.22
Explanation:
As you can see, your initial concentration of sodium acetate (NaCH₃COO) is 0.02M (0.02mol /L). When you add HCl, the reaction is:
NaCH₃COO + HCl → CH₃COOH + NaCl.
If you add HCl, and final concentration of NaCH₃COO is 0.015M, the concentration of CH₃COOH is 0.005M.
You can know the pH of this solution using H-H equation:
pH = pKa + log {NaCH₃COO} / {CH₃COOH}
pH = 4.74 + log {0.015M} / {0.005M}
pH = 5.22Select the correct answer from each drop-down menu.
Pictures that we receive from space are of the
✓ because it takes time for
to reach Earth.
Answer:
Pictures that we receive from space are of the past because it takes time for light to reach Earth.
Explanation:
For example, Mars is so far away that, depending on its position in orbit, a picture from Mars takes between 4 min and 24 min to reach Earth.
Answer: Pictures that we receive from space are of the
past
because it takes time for
light
to reach Earth.
Explanation:
When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is ____________.
When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is Hund's rule.
Hund's rules state that:
Every orbital in a sublevel has to be singularly occupied before any other orbital is able to be doubly occupied.
All of the electrons in single occupied orbitals have to have the same spin to maximize the total spin.
Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minute per gram of sample. Living wood gives 6.68 disintegrations per minute per gram. Estimate the date of occupation and hence the probable date of the wall painting in the Lascaux Cave. Hint: Disintegrations per minute per gram" has the same units as the time-derivative of concentration for a radioactive decay model. (You may use the fact that the half-life of ^14C is 5568 years.)
Answer:
Explanation:
count given by old sample = .97 disintegrations per minute per gram
count given by fresh sample = 6.68 disintegrations per minute per gram
Half life of radioactive carbon = 5568 years
rate of disintegration
dN / dt = λ N
In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .
Let initial no of radioactive be N₀ and after time t , number reduces to N
N₀ / N = 6.68 / .97
Now
[tex]N=N_0e^{-\lambda t}[/tex]
[tex]\frac{N}{N_0} =e^{-\lambda t}[/tex]
[tex]\frac{6.68}{.97} = e^{\lambda t}[/tex]
λ is disintegration constant
λ = .693 / half life
= .693 / 5568
= .00012446 year⁻¹
Putting the values in the equation above
[tex]\frac{6.68}{.97} = e^{.00012446\times t}[/tex]
[tex]6.8866 = e^{.00012446\times t}[/tex]
1.929577 = .00012446 t
t = 15503.6 years .
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
Answer:
19.07 g mol^-1
Explanation:
The computation of the molecular mass of the unknown gas is shown below:
As we know that
[tex]\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }[/tex]
where,
Diffusion rate of unknown gas = 155 mL/s
CO_2 diffusion rate = 102 mL/s
CO_2 molar mass = 44 g mol^-1
Unknown gas molercualr mass = M_unknown
Now placing these values to the above formula
[tex]\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}[/tex]
After solving this, the molecular mass of the unknown gas is
= 19.07 g mol^-1
Balance this equation: __ UO2(s) + __ HF(ℓ) → __ UF4(s) + __ H2O(ℓ) Though you would not normally do so, enter the coefficient of "1" if needed. UO2(s) HF(ℓ) UF4(s) H2O(ℓ)
Answer:
The balanced equation is given below:
UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)
The coefficients are: 1, 4, 1, 2
Explanation:
UO2(s) + HF(l) —› UF4(s) + H2O(l)
The above equation can be balance as follow:
There are 2 atoms of O on the left side and 1 atom on the right side. It can be balanced by writing 2 in front of H2O as shown below:
UO2(s) + HF(l) —› UF4(s) + 2H2O(l)
There are 4 atoms of F on the right side and 1 atom on the left side. It can be balanced by writing 4 in front of HF as shown below:
UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)
Now, the equation is balanced.
The coefficients are: 1, 4, 1, 2.
g For the following reaction, 20.9 grams of iron are allowed to react with 9.19 grams of oxygen gas . iron(s) oxygen(g) iron(II) oxide(s) What is the maximum mass of iron(II) oxide that can be formed
Answer:
26.87g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2Fe + O2 —> 2FeO
Next, we shall determine the masses of Fe and O2 that reacted and the mass of FeO produced from the balanced equation.
This is illustrated below:
Molar mass of Fe =56 g/mol
Mass of Fe from the balanced equation = 2 x 56 = 112 g
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32 g
Molar mass of FeO = 56 + 16 =72 g/mol
Mass of FeO from the balanced equation = 2 x 72 = 144 g
From the balanced equation above,
112 g of Fe reacted with 32 g of O2 to produce 144 g of FeO.
Next, we shall determine the limiting reactant.
This is illustrated below:
From the balanced equation above,
112 g of Fe reacted with 32 g of O2.
Therefore, 20.9 g of Fe will react with = (20.9 x 32)/112 = 5.97 g of O2.
From the calculations made above, we can see that only 5.97 g out of 9.19 g of O2 given were required to react completely with 20.9 g of Fe.
Therefore, Fe is the limiting reactant and O2 is the excess reactant.
Finally, we shall determine the mass of FeO produced from the reaction.
In this case, the limiting reactant will be used, as it will give the maximum yield of the reaction since all of it is used up in the reaction.
The limiting reactant is Fe and the maximum mass of FeO produced can be obtained as follow:
From the balanced equation above,
112 g of Fe reacted to produce 144 g of FeO.
Therefore, 20.9 g of Fe will react to produce = (20.9 x 144)/112 = 26.87g of FeO.
Therefore, the maximum mass of iron(II) oxide, FeO produced is 26.87g.
Identify the energy transformations in the following actions.
a. Turning on a space heater
b. Dropping an apple core into the garbage
c. Climbing up a rope ladder
d. Starting a car
e. Turning on a flashlight
Think about each action as a before and after. For instance, before the space heater is turned on, what kind of energy is there? After the space heater is turned on, what happens? This change from before to after will help you identify the two types of energy involved.
Identify five more types of energy transformations that you see at home, at school, or outdoors. Make sure to name the action, such as turning on a light, as well as the two types of energy involved. Remember, for energy to be transformed, the type of energy before and after a task must be different.
Answer:
to. Turn on a heater (electrical energy that is transformed into heat energy)
yes. Throwing an apple core in the trash (chemical energy that is transformed into nutritional energy for decomposers)
C. Climbing a rope ladder (chemical energy from our food into mechanical energy that allows us to climb the ladder)
re. Starting a car (electrochemical energy in mechanical energy)
me. Turning on a flashlight (chemical energy from the battery into light energy)
Five daily actions that exemplify the transformation of energy:
Do physical activity (change of metabolic energy in mechanics)
Turn on a heater (electrical energy that is transformed into heat energy)
Lighting a stove (Chemical energy product of combustion that results in the transformation of that energy into heat)
Turn on a cell phone (chemical energy characteristic of the battery that is transformed into sound and light energy)
Riding a bicycle (nutritional energy or energy of the metabolism that is transformed into mechanical energy)
Explanation:
The energy is never lost, it is always transformed.
It is one of the great principles of physics and is the reason why the universe is infinite.
What is the energy change associated with 1.5 mole of D being formed? 
Answer:
–36 KJ.
Explanation:
The equation for the reaction is given below:
2B + C —› D + E. ΔH = – 24 KJ
From the equation above,
1 mole of D required – 24 KJ of energy.
Now, we shall determine the energy change associated with 1.5 moles of D.
This can be obtained as illustrated below:
From the equation above,
1 mole of D required – 24 KJ of energy
Therefore,
1.5 moles of D will require = 1.5 × – 24 = –36 KJ.
Therefore, –36 KJ of energy is associated with 1.5 moles of D.
A volume of 30.0 mL of a 0.140 M HNO3 solution is titrated with 0.570 M KOH. Calculate the volume of KOH required to reach the equivalence point.
Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M [tex]*[/tex] 30 / 1000 L = 0.140 M [tex]*[/tex] 0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2 [tex]*[/tex] ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol [tex]*[/tex] ( 1 L / 0.570 mol ) [tex]*[/tex] ( 1000 mL / 1 L ) = 7.37 mL of KOH
Using the Ideal Gas Law, PV = nRT and the information below, solve the question. 5.01 moles of chlorine are held in a vessel with a fixed volume of 70L. What is the pressure of the gas in atm, if its temperature is 303K. REMEMBER TO USE R=0.08206L*atm / k*mol
Answer:
1.7795 atmExplanation:
[tex]n = 5.01\\V = 70L\\P = ?\\T = 303K\\R = 0.08206\\\\PV =nRT\\Make \:P \:Subject\:of\:the\:formula\\P = \frac{nRT}{V} \\\\P = \frac{5.01\times0.08206\times303}{70} \\\\P =\frac{124.5695}{70}\\ P = 1.7795[/tex]
g Increasing the number of unsaturations in a fatty acid ____________ the melting temperature of the fatty acid.
Answer:
Decreases
Explanation:
Fatty acid which have the double bond or triple bond are called unsaturated fatty acids. Because of the double or triple bond, unsaturated fatty acids are loosely packed and form some distance among molecules which lowers the melting point of unsaturated fatty acids.
So, if the unsaturation of fatty acid will increase, it leads to more branched and loosely packed molecules and decreases the melting temperature accordingly.