A student created a ball and stick molecule during the lesson on molecules and atoms. Examine the model shown here. Which statem

Answer:

The mole fraction of nitrogen is   0.52

Explanation:

Provided information:

Temperature =  31.2 °C

Pressure = 870.2 mmHg

Volume = 15.1 L

Mass of the mixture = 24.1 g

Mole fraction of nitrogen =?

Process:

Converting pressure:

870.2 / 760 = 1.12 atm

Converting temperature:

31.2 + 273 = 304.2 K

Total moles calculation:

PV = nRT

n = PV/RT

n =  1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K

n = 16.9 L.atm.  / 25 L.atm. mol⁻¹

n = 0.676 mol

Let x be the number of moles of nitrogen.

Thus, the moles of CO₂ = 0.676 - x

The mass of nitrogen = x mol × 28 g/mol, and for CO₂ the mass = 44 g/mol (0.676 - x)

We have 24.1  = 28x + (29.7 - 44x)

Rearranging gives: 24.1 - 29.7  =  28x  - 44x

-5.6 = -16 x

Therefore, x = 0.35

Mole fraction of nitrogen:

Mole fraction of nitrogen = moles of nitrogen / total moles

Mole fraction of nitrogen =   0.35  mol / 0.676 mol

Mole fraction of nitrogen =   0.52

A heavier player collides with a lighter player using greater force.

The lighter player sustains more injuries following the impact.

Explanation:

A heavier player impacts a lighter player with greater intensity, resulting in more pronounced injuries to the lighter player post-collision.

Force is defined as mass multiplied by the acceleration of an object;

    Force = mass x acceleration

We observe that as mass and acceleration increase, the force exerted rises accordingly.

Clearly, the heavier player's mass surpasses that of the lighter player, leading to a greater force exerted upon collision.

Moreover, the lighter player is likely to be injured more severely after the clash. The momentum generated by the heavier player during the impact is considerably significant. Once they collide, the lighter player will certainly alter their speed and trajectory.

Learn more:

Momentum

The true statement is B. With identical masses for both metals, the final temperature of the two will be more aligned with 498 K rather than 298 K, as iron's specific heat capacity is significantly greater than that of gold's.

Answer:

Angle ABE measures 27°.

Explanation:

Refer to the attached diagram related to this question.

The given values are ∠ABE=2n+7 and ∠EBF=4n-13.

Clearly seen in the diagram, ∠ABE and ∠EBF are equal in measure.

Move variable components to one side of the equation.

Split both sides by 2.

The solution for n arrives at 10.

The next step is to calculate ∠ABE.

Consequently, the measurement of angle ABE is 27°.

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

The half-reactions are:

Anode oxidation half reaction:  

Cathode reduction half reaction:  

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature =

n = electrons exchanged in oxidation-reduction = 2

= standard electrode potential for the cell = +0.63 V

= cell potential for the reaction =?

= concentration of Zn²⁺ = 3.5 M

=

Now substituting all known values into the equation, we arrive at:

So, the resulting cell potential for this reaction is 0.50 V