Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
What does the period number tell about the energy levels occupied by
electrons in an atom?
A. The period number tells how many electrons are in the highest
energy level of the atom.
B. The period number tells which is the highest energy level occupied
by the electrons.
C. The period number tells how many electrons are in each sublevel
of the atom.
D. The period number tells how many energy sublevels are occupied
in the atom.
Answer: B. The period number tells which is the highest energy level occupied by the electrons
Explanation:
The period number ( denoted by 'n' ) is the outer energy level that is occupied by electrons in an atom. The period number that an element is in, is the number of energy levels that the element has.When we move across a period from left to right in a periodic table the number of electrons in atoms increases within the same orbit.Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.
hence, the correct option is B. The period number tells which is the highest energy level occupied by the electrons.
The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.
What are energy levels ?The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.
When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.
Therefore, option B , second option is correct.
Learn more about energy levels at;
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plez hurry Which is an important safety precaution that should be taken during a tornado? Stay away from doors and windows. Move to high ground to avoid flood waters. Try to avoid the storm by driving or running. Stay outside to avoid being trapped in a building.
Answer: stay away from doors and windows.
Explanation:
to aviod geting hit by glass
Answer:
Stay away from doors and windows.
Explanation:
Always stay in the center of the room during a tornado storm. Avoid windows, doors, and corners. If you’re near a window, the glass can shatter and hurt you.
Diluting sulfuric acid with water is highly exothermic:
(Use data from the Appendix to find for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 MH2SO4(aq) (d = 1.060 g/mL). )
Suppose you carry out the dilution in a calorimeter. The initial T is 25.2°C, and the specific heat capacity of the final solution is 3.458 J/gK. What is the final T in °C ?
Answer:
The correct answer is 51.2 degree C.
Explanation:
The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.
Now the dHreaction = dHf (product) - dHf (reactant)
= -909.3 - (-814)
dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.
The heat change in calorimetry can be determined by using the formula,
q = mass * specific heat capacity * change in temperature -----------(i)
Based on the given information, the density of H₂SO₄ is 1.060 g/ml
The volume of H₂SO₄ is 1 Liter
Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams
The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.
ΔT = T₂ -T₁ = T₂ - 298.2 K
Now putting the values in equation (i) we get,
95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)
(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K
T₂ = 298.2 K + 26 K
T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.
what is the osmotic pressure of pure water
Answer:
The osmotic pressure of ocean water is about 27 atm.
Explanation:
Pure water is water that contains no impurities. Ocean water is 96.5% pure with only about 3.5% of its content, salt water.
Osmotic pressure occurs when solutions that have different concentrations are isolated by a membrane. This osmotic pressure makes water move towards the solution that has the highest concentration, which means that if the concentration or temperature of the solution is high, the osmotic pressure becomes higher.
The equation for osmotic pressure is pi = iMRT.
Based on their molecular structure, identify the stronger acid from each pair of oxyacids. Match the words in the left column to the appropriate blanks in the sentences on the right.
1) HI is a stronger acid than H2Te because iodine____than tellurium.
2) H2Te is a stronger acid than H2S because the H-Te bond is_____.
3) NaH is not acidic because hydrogen____than sodium.
a. has a more negative electron afflity
b. is more electronegative
c. has a larger atomic radius
d. stronger
e. is harder to ionize
Answer:
1)is more electronegative
2)
3) is more electronegative
Explanation:
1) for the first question, iodine is more electronegative than tellurium hence we naturally expect that HI should be more acidic than H2Te since electronegativities play a role in the acidity of chemical species.
2) the correct option is not listed because the H2Te bond is weaker than the H2S bond. This makes it easier for H2Te to dissociate releasing H^+ , thereby being more acidic than H2S.
3) Hydrogen is more electronegative than sodium hence it cannot be ionized thus NaH is not acidic.
Choose the substance with the lowest boiling point.
A. NBr3.
B. CI2H2.
C. H2O2.
D. H2S.
E. O2.
Answer:
E. O2
Explanation:
All substances has a simple molecular structure, where between their molecules are held by van der Waals' forces. But C must be incorrect because between the H2O2 molecules, they are mainly held by hydrogen bonds on top of van der Waals' forces. Hydrogen bonds are stronger than van der Waals' forces, so more energy is required to separate the H2O2 molecules.
In structures A and D, the molecules are polar. Their van der Waals' forces are stronger than Cl2H2 and O2, which are non-polar.
Between the Cl2H2 and O2, O2 has a smaller molecular size. The van der Waals' forces between the O2 molecules are hence the weakest. Least amount of energy is required to break the intermolecular forces between the O2 molecules therefore it has the lowest boiling point.
What is the name of this molecule?
Answer:
[tex]\boxed{Butyne}[/tex]
Explanation:
Triple Bonds => So it is an alkyne
The suffix used will be "-yne"
4 Carbons => The prefix used will be "But-"
Combining the prefix and suffix, we get:
=> Butyne
Answer:
[tex]\boxed{\mathrm{Butyne}}[/tex]
Explanation:
Alkynes have triple bonds ≡. The molecule has one triple bond.
Suffix ⇒ yne
The molecule has 4 carbon atoms and 6 hydrogen atoms.
Prefix ⇒ But (4 carbons)
The molecule is Butyne.
[tex]\mathrm{C_4H_6}[/tex]
Which of the following statements about water is not true?
Answer:
Water has a low specific heat capacity and so large bodies of water moderate temperatures on Earth.
Explanation:
Water has a very high specific heat capacity, meaning that it has to absorb a lot of energy to raise the temperature by one degree. Because water has a high specific heat capacity, large bodies of water can moderate the temperature of nearby land.
Hope this helps.
Fructose-2,6-bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose-1,6-bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose-2,6-bisphosphate is regulated by many hormones, second messengers, and enzymes.
How do the following affect glycolysis and gluconeogenesis?
Activate glycolysis Inhibit gluconeogenesis Activate gluconeogenesis Inhibit glycolysis
1. increased levels of fructose-2,6-bisphosphatase
2. activation of fructose-2,6-bisphosphate (FBPase-2)
3. increased glucagon levels
4. activation of PFK-2
5. increased levels of CAMP
Answer:
1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis
2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis
3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis
4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis
5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis
Explanation:
Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.
Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.
Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.
The following shows the various regulatory methods and their effects on both processes:
1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.
Increased levels of fructose-2,6-bisphosphatase activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.
2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.
3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.
4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.
5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the activation of gluconeogenesis and inhibition of glycolysis.
If the average rate of the reaction A --->2B C is 1M/s, what is the average rate of formation (in M/s) of B over that same period of time
Answer:
[tex]r_B=2M/s[/tex]
Explanation:
Hello,
In this case, since the average rate of reaction is related with the consumption of A which has an stoichiometric coefficient of 1, the rate of formation of B will be:
[tex]r_B=2*1M/s\\\\r_B=2M/s[/tex]
By cause of the stoichiometric coefficient of B which doubles the average rate.
Best regards.
Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.
Answer:
c iodine
Explanation:
fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride
Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you
a) H2(g)+I2(s)--->2HI(g)
b) MnO2(s)+2CO(g)--->Mn(s)+2CO2(g)
c) NH4Cl(s)--->NH3(g)+HCl(g)
Answer:
a) [tex]\Delta G=2.6kJ[/tex]
b) [tex]\Delta G=-979.57kJ[/tex]
c) [tex]\Delta G=264.21kJ[/tex]
Explanation:
Hello,
In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:
[tex]\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol[/tex]
So we proceed as follows:
a)
[tex]\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ[/tex]
b)
[tex]\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ[/tex]
c)
[tex]\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ[/tex]
Regards.
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 chemical equation of Braium chloride?
Answer:
BaCl2
Explanation:
Barium = Ba
Chloride => Cl-
Chemical Equation:
Ba + Cl => BaCl2
Note:
The valency of barium is 2 and valency of chloride is 1 (i.e. chlorine). The formula formed by the combination of these elements is BaCl2 (there's exchange of valencies when these two elements combine).
In the laboratory you are asked to make a 0.694 m copper(II) iodide solution using 455 grams of water. How many grams of copper(II) iodide should you add
Answer:
100.2g of CuI₂ you must add
Explanation:
Molality, m, is defined as the ratio between moles of solute and kg of solvent.
In the problem, you have a 0.694m of copper (II) iodide -CuI₂, molar mass: 317.35 g/mol-. That means there are 0.694 moles of CuI₂ per kg of water.
As you have 455g = 0.455kg of water -solvent-, moles of CuI₂ are:
0.455kg ₓ (0.694 moles CuI₂ / kg) = 0.316 moles of CuI₂
Using molar mass, grams of CuI₂ in the solution are:
0.316moles CuI₂ ₓ (317.35g / mol) =
100.2g of CuI₂ you must addWrite electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For example, the electron configuration of LiLi would be entered in complete form as 1s^22s^1 or in condensed form as [He]2s^1.
Answer:
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Explanation:
Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),
Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of 48 electrons surrounding its nucleus.
The electronic configuration of Cadmium is;
Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
The shorthand notation is given as;
Cd: [Kr] 4d¹⁰5s²
Cd2+ means that it has two less electrons, hence it's electron configuration is given as;
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Which accurately describes one impact of the atmosphere on Earth’s cycles?
Answer:
Produces Wind Currents
Explanation:
Answer:
produces wind currents
Explanation:
i just took the test and got it right :}
A 40.80 gram sample of copper is heated in the presence of excess sulfur. A metal sulfide is formed with a mass of 51.09 g. Determine the empirical formula of the metal sulfide.
Answer:
Cu₂S
Explanation:
From the question,
Cu S
Mass: 40.80 g 51.09-40.80 = 10.29 g
Mole ratio: 40.80/63.5 10.29/32.1
0.64 : 0.32
Divide by the smallest,
0.64/0.32 : 0.32/0.32
2 : 1
Therefore,
Empirical formula = Cu₂S.
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo se elevo a una altitud donde la temperatura era de -20°C y la presión de 425 torr, ¿Cuál era el volumen del gas del globo en estas condiciones?
Answer:
El volumen del gas era 12.95 L
Explanation:
Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:
“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”
La ley de Boyle se expresa matemáticamente como: P*V=k
Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:
[tex]\frac{V}{T}=k[/tex]
Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:
[tex]\frac{P}{T}=k[/tex]
Combinado las mencionadas tres leyes se obtiene:
[tex]\frac{P*V}{T} =k[/tex]
Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:
P1: 750 torrV1: 8.5 LT1: 20°C= 293°K (siendo 0°C=273°K)P2: 425 torrV2: ?T2: -20°C= 253 °KReemplazando:
[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]
Resolviendo:
[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]
V2= 12.95 L
El volumen del gas era 12.95 L
A hot lump of 27.4 g of aluminum at an initial temperature of 69.5 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.
Answer:
[tex]\large \boxed{29.7 \,^{\circ}\text{C}}[/tex]
Explanation:
There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.
According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.
Let the Al be Component 1 and the H₂O be Component 2.
Data:
For the Al:
[tex]m_{1} =\text{27.4 g; }T_{i} = 69.5 ^{\circ}\text{C; }\\C_{1} = 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
For the water:
[tex]m_{2} =\text{50.0 g; }T_{i} = 25.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
Calculations
(a) The relative temperature changes
[tex]\begin{array}{rcl}\text{Heat lost by Al + heat gained by water} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{27.4 g}\times 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{50.0 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\24.74\Delta T_{1} + 209.2\Delta T_{2} & = & 0\\\end{array}[/tex]
(b) Final temperature
[tex]\Delta T_{1} = T_{\text{f}} - 69.5 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 25.0 ^{\circ}\text{C}[/tex]
[tex]\begin{array}{rcl}24.74(T_{\text{f}} - 69.5 \, ^{\circ}\text{C}) + 209.2(T_{\text{f}} - 25.0 \, ^{\circ}\text{C}) & = & 0\\24.74T_{\text{f}} - 1719 \, ^{\circ}\text{C} + 209.2T_{\text{f}} -5230 \, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} - 6949\, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} & = & 6949 \, ^{\circ}\text{C}\\T_{\text{f}}& = & \mathbf{29.7 \, ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature is $\large \boxed{\mathbf{29.7 \,^{\circ}}\textbf{C}}$}[/tex]
Check:
[tex]\begin{array}{rcl}27.4 \times 0.903 \times (29.7 - 69.5) + 50.0 \times 4.184 (29.7 - 25.0)& = & 0\\24.74(-39.8) +209.2(4.7) & = & 0\\-984.6 +983.2 & = & 0\\-985 +983 & = & 0\\0&=&0\end{array}[/tex]
The second term has only two significant figures because ΔT₂ has only two.
It agrees to two significant figures
Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compound xy can be generated 2x + y2 = 2xy
Answer:
[tex]4.36~g~XY[/tex]
Explanation:
In this case, we can start with the reaction:
[tex]2X + Y_2~->~2XY[/tex]
If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:
[tex]3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X[/tex]
[tex]4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2[/tex]
Now, we can divide by the coefficient of each compound (given by the balanced reaction):
[tex]\frac{0.04~mol~X}{1}=~0.04[/tex]
[tex]\frac{0.0875~mol~Y_2}{2}=0.04375[/tex]
The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:
[tex]0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY[/tex]
Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol [tex]Y_2[/tex] = 48 g [tex]Y_2[/tex] (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:
[tex]0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY[/tex]
I hope it helps!
The equilibrium between carbon dioxide gas and carbonic acid is very important in biology and environmental science. CO2 ( aq) + H2O ( l) H2CO3 ( aq) Which one of the following is the correct equilibrium constant expression (K c) for this reaction?
a) K =[H2CO3]/ [CO2]
b) K=[CO2]/ [H2CO3]
c) K=[H2CO3]/ [CO2][H2O]
d) K=[CO2][H2O]/ [H2CO3]
e) K=1/[H2CO3]
Answer:
Kc = [H₂CO₃] / [CO₂]
Explanation:
Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.
Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.
Also, each specie must be powered to its reactant coefficient.
For example, for the reaction:
aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)
The equilibrium constant, kc is:
Kc = [D]ⁿ / [B]ᵇ[E]ˣ
You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants
Thus, for the reaction:
CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)
The Kc is:
Kc = [H₂CO₃] / [CO₂]
Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K if the pressure of each gas is 22.20 mm Hg.
Answer:
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
ΔG ≅ 199.91 kJ
Explanation:
Consider the reaction:
[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]
temperature = 298.15K
pressure = 22.20 mmHg
From, The standard Thermodynamic Tables; the following data were obtained
[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :
[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]
[tex]K_p = \dfrac{1}{ (22.20)}[/tex]
[tex]K_p[/tex] = 0.045
[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]
where;
R = gas constant = 8.314 × 10⁻³ kJ
[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]
[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]
[tex]\Delta G =207.6+ (-7.687048037)[/tex]
[tex]\Delta G =[/tex] 199.912952 kJ
ΔG ≅ 199.91 kJ
A gas mixture containing N2 and O2 was kept inside a 2.00 L container at a temperature of 23.0°C and a total pressure of 1.00 ATM the partial pressure of oxygen was 0.722 ATM how many grams of nitrogen are present in the gas mixture
Answer:
0.641 g of Nitrogen are present in the mixture.
Explanation:
We use the Ideal Gases Law, to solve this question.
For the mixture:
P mixture . V mixture = mol mixture . R . T
We convert the T° to K → 23°C + 273 = 296 K
R = Ideal gases constant → 0.082 L.atm/mol.K
1 atm . 2L = mol mixture . 0.082 L.atm/mol.K . 296K
2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles
We know that sum of partial pressure = 1
Partial pressure N₂ + Partial pressure O₂ = 1
1 - 0.722 atm = Partial pressure N₂ → 0.278 atm
We apply the mole fraction concept:
Partial pressure N₂ / Total pressure = Moles N₂ / Total moles
Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles
Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles
We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g
641 mg
3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard
Answer:
High purity.
Stability (low reactivity)
Low hygroscopicity (to minimize weight changes due to humidity)
Explanation:
There are different primary standards that could be used in a standardization titration in order to achieve the best and accurate result possible. These standards include high purity,stability and low hygoscropicity .
A high purity means the reactants lack impurities which could affect the result. Stability also ensures that there is non reactivity with elements/compounds in the atmosphere while low hygroscopicity ensures weight changes are minimized due to humidity.
Which is the electron configuration for bromine?
Answer:
The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.
Explanation:
1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^5
Which correctly lists the three land uses that the Bureau of Land Management was originally created to manage? mining, recreation, wildlife refuges recreation, developing oil and gas, battlefields grazing, mining, developing oil and gas developing oil and gas, battlefields, wildlife refuges
Answer: C
Explanation:
Right on edge 2020
The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.
What is land management?Land management refers to the activities which are done in order to protect and preserve the land as well the resources found on land.
The Bureau of Land Management was created to manage land in the US.
The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.
Learn more about land management at: https://brainly.com/question/784519
A spinning turbine can generate electricity only in the form of a/an _______ current.
Of all the alternative energy technologies presented in this section, only solar panels produce a/an _______ current.
Answer:
The correct answer is - alternating and direct, in order.
Explanation:
Alternating current is is type of electric current that is characterized by the direction of the flow of electrons in continuously switches its directs in opposite manner at regular cycles. While direct current or DC is flow of the electrons that move from starting to end in one direction.
Spinning turbines always leads to the alternating electric current while only solar energy produces the direct current with the help of the solar panels.
Thus, the correct answer is - alternating and direct, in order.
Answer:
1. alternating
2. direct
3. The sun heats up the atmosphere as Earth spins, creating areas of high and low temperature. This temperature difference causes wind to start moving through convection, which can then drive a wind turbine to produce electricity.
Explanation:
From Penn
identify the correct acid/conjugate base pair in this equation:
NaHCO3 + H20 = + H2CO3 + OH
+ Na
H20 is an acid and H2CO3 is its conjugate base.
HCO3 is an acid and OH is its conjugate base.
H20 is an acid and HCO3 is its conjugate base.
H20 is an acid and OH is its conjugate base.
Answer:
H20 is an acid and OH is its conjugate base.
Explanation:
Chemical reactions involving acids and bases occur. An acid is a substance that dissociates in water i.e. lose an hydrogen ion/proton. According to the Bronsted-Lowry acid-base theory, when an acid dissociates in water and loses its hydrogen ion, the resulting substance that forms is the CONJUGATE BASE. A conjugate base is the compound formed as a result of the removal of an H+ ion from an acid.
Based on the chemical reaction in the question, NaHCO3 + H20 = H2CO3 + OH- + Na+
The H20 loses its hydrogen ion (H+) to form an anion OH-. This anion formed is the conjugate base while H20 is its acid.
A. Identify the structure drawn below.
Answer:
Hexane
Explanation:
You have a carbon structure with only single bonds. This means that the name will end in -ane.
There are 6 carbon atoms. This means that the name will begin with hex-.
The structure is hexane.