The electronegativity affects the polarity of the bond between two atoms, as the more electronegative atom will make its end of the bond more negative. The correct option is C.
Electronegativity is a charge that shows the ability of an element to gain electron pairs with other elements during bonding. Electronegativity is altered by the distance between the electron and the nuclei and the atomic number of the element.
Polarity is the state of the atomic body in which it has placed charges in an opposite way to the other atoms so that they can join together.
Thus, the correct option is C. The more electronegative atom will make its end of the bond more negative.
To learn more about electronegativity, refer to the link:
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Answer:
C. The more electronegative atom willl make its end of the
bond more negative
A P E X
Answer:
Non polar covlant
Explanation:
Answer:
Partial pressure of neon = 175 mmHg
Partial pressure of xenon = 564 mmHg
Explanation:
The partial pressure of a gas in a mixture can be calculated as the product of the mole fraction of the gas (Xi) and the total pressure (Pt), as follows:
Pi = Xi Pt
The total pressure is 739 mmHg ⇒ Pt = 739 mmHg
In order to calculate the mole fraction of each gas, we have to first calculate the number of moles of each gas (n) by dividing the mass of the gas into the molar mass (MM):
For neon gas (Ne):
MM(Ne) = 20.18 g/mol
n(Ne)= mass/MM = 0.919 g x 1 mol/20.18 g = 0.045 mol Ne
For xenon gas (Xe):
MM(Xe) = 131.3 g/mol
n(Xe)= mass/MM = 19.1 g x 1 mol/131.3 g = 0.145 mol Xe
Now, we calculate the mole fraction (X) by dividing the number of moles of the gas into the total number of moles (nt):
nt= moles Ne + moles Xe = 0.045 mol + 0.145 mol = 0.190 mol
X(Ne) = moles Ne/nt = 0.045 mol/0.190 mol = 0.237
X(Xe) = moles Xe/nt = 0.145/0.190 mol = 0.763
Finally, we calculate the partial pressures of Ne and Xe as follows:
P(Ne) = X(Ne) x Pt = 0.237 x 739 mmHg = 175 mmHg
P(Xe) = X(Xe) x Pt = 0.763 x 739 mmHg = 564 mmHg
Actually Welcome to the Concept of the Kinematics.
Here, we know that, Velocity = Distance / Time,
So here, Distance = 10km = 10×1000 = 10000 metres.
, Time = 14 min 30 sec = 870 seconds,
so now, we get velocity as,
=> V = 10000 ÷ 870 => 11.49 m/s .
Hence, Velocity is 11.49 m/s.
Answer:
1.133 kPa is the average pressure exerted by the molecules on the walls of the container.
Explanation:
Side of the cubic box = s = 20.0 cm
Volume of the box ,V=
Root mean square speed of the of helium molecule : 200m/s
The formula used for root mean square speed is:
where,
= root mean square speed
k = Boltzmann’s constant =
T = temperature = 370 K
M = mass helium =
= Avogadro’s number =
Moles of helium gas = n
Number of helium molecules = N =
N =
Ideal gas equation:
PV = nRT
Substitution of values of T and n from above :
(1 Pa = 0.001 kPa)
1.133 kPa is the average pressure exerted by the molecules on the walls of the container.
The question asks for the average pressure exerted by helium gas molecules on the walls of a cubic container. Using the equation PV = Nmv^2, we can calculate pressure by substituting the given values for volume, number of molecules, mass of one molecule, and root-mean-square speed.
The question is asking to calculate the average pressure exerted by helium gas molecules on the walls of a cubic container. The important formula relating pressure (P), volume (V), number of molecules (N), mass of a molecule (m), and the square of the rms speed (v2) of the molecules in a gas is:
PV = Nmv2,
First, we need to determine the volume of the container, which is the cube of one side, so V = (20 cm)3 = (0.2 m)3. Inserting the given values into the equation and solving for P gives us the desired answer. Recall that the rms speed is given, so no temperature calculations are needed.
Therefore, using all given data points:
Volume (V) = (0.2 m)3
Number of molecules (N) = 2.00 × 1023
Mass of one helium molecule (m) = 3.40 × 10-27 kg
Root-mean-square speed (vrms) = 200 m/s
By substituting these values, we can find the pressure exerted by the gas. This represents an application of kinetic theory of gases which assumes the behavior of an ideal gas.
Answer:
Kc = 2.145 × 10⁻⁸¹
Explanation:
Let's consider the following reaction:
O₂(g) ⇄ 2O(g)
The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:
ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)
where,
ni are the moles of products and reactants
ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants
In this case,
ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)
ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol
ΔG° = 460.2 kJ/mol
With this information, we can calculate the equilibrium constant (Kc) using the following expression: