Answers
Answer:
The correct answer is Pu, 234.
Explanation:
In the given case, let us consider the reactant as X. Now the mass number (balanced) on both the sides will be,
Mass of X = Mass of Molybdenum + Mass of Tin + Mass of neutrons
M = 1 * 103 + 1 * 131 + 2 * 0
M = 234
Now the atomic number (balanced) on both the sides,
Atomic number of X = Atomic number of Molybdenum + Atomic number of Tin + Atomic number of neutrons
A = 1*42 + 1*50 + 2*1
A = 94
The atomic number 94 is for the element Plutonium, whose symbol is Pu. Thus, the reactant is 234-Pu.
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14) Describe the Cloud Model.
The higher the pH, the less acidic the solution
Is 3.5:Aqueous solutions of iron(III) bromide and ammonium carbonate react to form a precipitate. Answer the follwingquestions with regards to this reaction.a) Write the molecular equation for this reaction byTranslating the two reactants into their chemical formulae.Predict the products.Label all the states.Balance the reaction.
Answers
Answers
Answer:
0.2320V
Explanation:
Voltage can be defined as the amount of potential energy available (work to be done) per unit charge, to move charges through a conductor.
Voltage can be generated by means other than rubbing certain types of materials against each other.
Please look at attached file for solution to the problem.
Final answer:
The expected voltage generated by this concentration cell is approximately 0.113 V.
Explanation:
To calculate the voltage generated by the concentration cell, we can use the Nernst equation. The Nernst equation relates the concentration of the ions in the two compartments to the voltage of the cell. The equation is:
E = E° - (RT/nF) ln(Q)
Where:
- E is the voltage of the cell
- E° is the standard cell potential
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin (25 + 273 = 298 K)
- n is the number of moles of electrons transferred (2 in this case)
- F is Faraday's constant (96,485 C/mol)
- ln(Q) is the natural logarithm of the reaction quotient
The reaction quotient (Q) can be calculated using the concentrations of the lead (II) and iodide ions in each compartment.
Since this is a concentration cell, the standard cell potential (E°) for this system is 0 V. Therefore, the equation simplifies to:
E = - (RT/nF) ln(Q)
Now we can calculate the voltage:
- Calculate Q:
The solubility product constant (Ksp) for PbI2 is 1.4 x 10-8. Because PbI2 is in a saturated solution, the concentration of Pb2+ ions and I- ions are both equal to the solubility of PbI2. We can substitute these values into the equation to calculate Q:
Q = [Pb²+] x [I-]
Q = (1.4 x 10-8) x (1.4 x 10-8) = 1.96 x 10-16
- Calculate E:
Now we can calculate the voltage using the given values:
For the Nernst equation, we need to convert the temperature to Kelvin:
T = 25°C + 273 = 298 K
Substitute the values into the equation:
E = - (8.314 J/mol·K x 298 K / 2 x 96,485 C/mol) ln(1.96 x 10-16)
E ≈ 0.113 V
Therefore, the expected voltage generated by this concentration cell is approximately 0.113 V.
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Be
Ca
Ba
Sr
Answers
Answer:
none
Explanation:
it's Fr. which is francium.
Explanation:
An atomic radius is defined to be one-half the distance between the nuclei of two atoms, assuming a spherical atom since, according to the quantum mechanical model of the atom, electrons are located within a probability cloud surrounding the nucleus which has no sharp boundary.
Notice that, in general, there are two main trends of atomic radii in the Periodic Table of Elements.
The first trend illustrates that atomic radii increase when going down a group in the periodic table. This is because when moving downwards in a group, every subsequent atom gains an additional principal energy level, which leads to electron shielding. Electron shielding refers to the decreased attraction between the electrons that occupy the higher principal energy level and the nucleus of the atom due to the shielding of electrons in the lower principal energy level.
The second trend outlines that atomic radii decrease when going across the period from left to right. For elements within a period, individual electrons occupy the same principal energy level. Likewise, when an electron is added, a new proton is also added to the nucleus, providing the nucleus with a stronger positive charge and hence leading to a higher effective nuclear charge. This increase in nuclear attraction pulls the electrons closer towards the nucleus, leading to a decrease in atomic radius.
Therefore, given the option between beryllium, calcium, barium, and strontium, the element with the largest atomic radius is barium since all the elements given are in Group II, however, barium is the element furthest down the group and therefore have electrons occupying the highest principal energy level compared to other elements.
Answers
C3H8(g) + 5O2(g) ↔ 3CO2(g) + 4 H2O (g)
aA + bB ↔ cC + dD
according to K formula when:
K = concentration of the products / concentration of the reactants
K = [C]^c[D]^d / [A]^a[B]^b
when [A],[B],[C]and[D] is the concentrations
and a,b,c and d is the no of moles
∴ K = [CO2]^3[H2O]^4 / [C3H8] [ O2]^5
Answer:
combustion reaction
Explanation:
there is oxides in the equation c3h8(g)+5o2(g)3co2(g)+4h2o(g)
Answers
Explanation:
Molar mass of potassium nitrate will be calculated as follows.
Molar mass = molar mass of K + molar mass of N + 3 × molar mass of O
= 39.098 g/mol + 14.006 g/mol + 3 × 15.999 g/mol
= 102.102 g/mol
Now, adding the given amount of potassium nitrate present in each beaker as follows.
(2.3 + 1.91 + 5.985 + 0.52) g
= 10.715 g
Therefore, calculate number of moles as follows.
No. of moles =
=
= 0.105 mol
Thus, we can conclude that 0.105 moles of potassium nitrate were recovered after the water evaporated.
Answers
Answer:
a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)
b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.
c) H⁺, HSO₄⁻, SO₄²⁻
d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.
f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
Explanation:
a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.
The balanced equation is:
H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l) [1]
b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?
In the beginning there is H₂SO₄ and the ions that come from its dissociation reactions: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its ionization.
H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
c) What is the conducting species in this initial solution?
The main responsible for conductivity are the ions coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.
d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?
As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?
At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are weak electrolytes, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.
f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?
After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
Final answer:
The chemical reaction between H2SO4 and Ba(OH)2 forms BaSO4 and water, reducing conductivity by reducing the number of free ions. Beyond the equivalence point, the conductivity increases due to the dissociated ions from the excess Ba(OH)2 in the solution.
Explanation:
Chemical Reaction and Metric Titration
Firstly, the equation representing the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:
Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)
In the beginning, the solution only contains H2SO4 with its dissociated ions serving as the conducting species. As titrant (Ba(OH)2) is added, they react to form BaSO4, a solid precipitate reducing the number of free ions in the solution, thus decreasing conductivity. At the equivalence point, all H2SO4 has reacted, and conductivity reaches its minimum as there are lesser free ions for conduction. If additional titrant is added past the equivalence point, conductivity increases due to excess Ba(OH)2's dissociated ions that increase ion concentration in solution.
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