Answer:
Option (A) saturated and is at equilibrium with the solid KCl
Explanation:
A saturated solution is a solution which can not dissolve more solute in the solution.
From the question given above, we can see that the solution is saturated as it can not further dissolve any more KCl as some KCl is still visible in the flask.
Equilibrium is attained in a chemical reaction when there is no observable change in the reaction system with time. Now, observing the question given we can see that there is no change in flask as some KCl is still visible even after thorough shaking. This simply implies that the solution is in equilibrium with the KCl solid as no further dissolution occurs.
Largest radius to Smallest radius
Answer Bank
K
Ca
Ga
Ge
As
Sc
Br
Kr
The elements according to the decreasing atomic radius are arranged as-
K, Ca, Sc, Ga, Ge, As, Br, Kr
An atomic radius is half the distance between adjacent atoms of the same element in a molecule. It is a measure of the size of the element’s atoms, which is typically the mean distance from the nucleus centre to the boundary of its surrounding shells of the electrons.
An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements. The atomic radius decreases on moving from left to right across a period.
Thus the elements according to the decreasing atomic radius are arranged as -
K, Ca, Sc, Ga, Ge, As, Br, Kr
Learn more about Atomic radius, here:
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Answer:
1a. The balanced equation is given below:
2NO + O2 → 2NO2
The coefficients are 2, 1, 2
1b. 755.32g of NO2
2a. The balanced equation is given below:
2C6H6 + 15O2 → 12CO2 + 6H2O
The coefficients are 2, 15, 12, 6
2b. 126.25g of CO2
Explanation:
1a. Step 1:
Equation for the reaction. This is given below:
NO + O2 → NO2
1a. Step 2:
Balancing the equation. This is illustrated below:
NO + O2 → NO2
There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:
2NO + O2 → 2NO2
The equation is balanced.
The coefficients are 2, 1, 2
1b. Step 1:
Determination of the limiting reactant. This is illustrated below:
2NO + O2 → 2NO2
From the balanced equation above, 2 moles of NO required 1 mole of O2.
Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.
From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.
1b. Step 2:
Determination of the maximum amount of NO2 produced. This is illustrated below:
2NO + O2 → 2NO2
From the balanced equation above, 2 moles of NO produced 2 moles of NO2.
Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.
1b. Step 3:
Conversion of 16.42 moles of NO2 to grams. This is illustrated below:
Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol
Mole of NO2 = 16.42 moles
Mass of NO2 =?
Mass = number of mole x molar Mass
Mass of NO2 = 16.42 x 46
Mass of NO2 = 755.32g
Therefore, the maximum amount of NO2 produced is 755.32g
2a. Step 1:
The equation for the reaction.
C6H6 + O2 → CO2 + H2O
2a. Step 2:
Balancing the equation:
C6H6 + O2 → CO2 + H2O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:
C6H6 + O2 → 6CO2 + H2O
There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:
C6H6 + O2 → 6CO2 + 3H2O
There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:
C6H6 + 15/2O2 → 6CO2 + 3H2O
Multiply through by 2 to clear the fraction.
2C6H6 + 15O2 → 12CO2 + 6H2O
Now, the equation is balanced.
The coefficients are 2, 15, 12, 6
2b. Step 1:
Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:
2C6H6 + 15O2 → 12CO2 + 6H2O
Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol
Mass of C6H6 from the balanced equation = 2 x 78 = 156g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 15 x 32 = 480g
2b. Step 2:
Determination of the limiting reactant. This is illustrated below:
From the balanced equation above,
156g of C6H6 required 480g of O2.
Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.
From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.
2b. Step 3:
Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb
2C6H6 + 15O2 → 12CO2 + 6H2O
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 12 x 44 = 528g
2b. Step 4:
Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:
From the balanced equation above,
156g of C6H6 produced 528g of CO2.
Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2
The given reaction will shift towards cis-2-butene once placed in equilibrium. This can be determined by calculating the reaction quotient and comparing it with the equilibrium constant.
The reaction could either shift towards the cis-2-butene or trans-2-butene depending on whether the reaction quotient, Q, is lesser or greater than the equilibrium constant, Kp.
Bear in mind that Kp = Ptrans/Pcis. Let's say that Pt is the partial pressure of trans-2-butene and Pc is the partial pressure of cis-2-butene at equilibrium. If we start with 5 atm of each gas, the change in Pc is -x and the change in Pt is +x.
So, Kp = (5+x)/(5-x). We are given that Kp = 3.4. Solving these two equations will show that x is a negative value, which means that the system shifts towards cis-2-butene.
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For the isomerization reaction cis-2-butene ⇌ trans-2-butene, with an initial pressure of 5.00 atm for both gases and a Kp of 3.40, the system will shift towards the product, trans-2-butene, as Kp > Qp (1). This reflects the principle that a chemical system at equilibrium will shift to counteract any change.
In terms of the equilibrium constant (K), for gas-phase reactions, Kp represents equilibrium in terms of partial pressures, while Kc represents it in molar concentrations. For instance, in the isomerization reaction given cis-2-butene ⇌ trans-2-butene, Kp is given as 3.40. To determine the behavior of the system, we need to compare it to reaction quotient (Q). Given that the flask initially contains 5.00 atm of each gas, Qp is 1 (since Qp = partial pressure of trans-2-butene / partial pressure of cis-2-butene). Since Kp > Qp, the reaction will shift towards the products, hence the system will shift towards trans-2-butene. From this, it is clear that the equilibrium constant and reaction quotient play vital roles in determining the direction of shift in a chemical equilibrium.
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Answer:
p orbitals only
Explanation:
Carbon has an atomic number of 6 so its electron configuration will be 1s² 2s² 2p². It has two orbitals as indicated with the 2 as its period number with the outer orbital have 4 valence electrons. So carbon is in the p-orbital, period 2 and in group 4.
Carbon's valence electrons reside in the 2s and 2p orbitals. These orbitals hybridize during bond formation to create equivalent sp3 hybrid orbitals, as evidenced in the methane molecule. Carbon's valence electrons are not placed in d orbitals.
Carbon (atomic number 6) has a total of six electrons. Two of these fill the 1s orbital. The next two fill the 2s orbital, and the final two are in the 2p subshell. According to Hund's rule, the most stable configuration for an atom is one with the maximum number of unpaired electrons. Therefore, carbon has two electrons in the 2s subshell and two unpaired electrons in two separate 2p orbitals. When discussing valence electrons, the electrons in the outermost shell are the ones considered, which for carbon are the electrons in the second shell namely 2s and 2p.
The geometry of the methane molecule (CH4) illustrates that in the bonding process, the s and p orbitals hybridize to allow the formation of four equivalent bonds with hydrogen atoms. Without hybridization, we would expect three bonds at right angles (from the p orbitals) and one at a different angle (from the s orbital). Nonetheless, through orbital hybridization, all four bonds in methane are identical, which is explained by the concept of sp3 hybridized orbitals.
Therefore, the valence electrons for carbon would be placed in the s orbital and p orbitals, not in the d orbitals, because carbon does not have electrons in the d subshell in its ground state. Additionally, the s and p orbitals are the only ones involved in bonding for carbon in most of its compounds, such as methane.
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