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2 years ago

Which statement correctly describes the relationship between thermal energy and particle movement?

1 answer:

5 0

As thermal energy increases, there is more particle movement. As thermal energy increases, there is more particle movement. As thermal energy increases, there is less particle movement.

Sure hope this helps you

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How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at

amid [387]

Kinetic energy increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the stopping distance should increase by a factor of four, assuming that the driver is can apply the brakes with sufficient precision to almost lock the brakes.

5 0

2 years ago

Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in

Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

$4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $X_2+3Y_2\rightarrow 2XY_3$ $\Delta H_1=-370kJ$

(2) $X_2+2Z_2\rightarrow 2XZ_2$ $\Delta H_2=-120kJ$

(3) $2Y_2+Z_2\rightarrow 2Y_2Z$ $\Delta H_3=-270kJ$

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) $4XY_3\rightarrow 2X_2+6Y_2$ $\Delta H_1=2\times (+370kJ)=740kJ$

(2) $2X_2+4Z_2\rightarrow 4XZ_2$ $\Delta H_2=2\times (-120kJ)=-240kJ$

(3) $6Y_2+3Z_2\rightarrow 6Y_2Z$ $\Delta H_3=3\times (-270kJ)=-810kJ$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+740kJ)+(-240kJ)+(-810kJ)$

$\Delta H=-310kJ$

Therefore, the change in enthalpy of the reaction is, -310 kJ

5 0

3 years ago

Consider a plywood square mounted on an axis that is perpendicular to the plane of the square and passes through the center of t

belka [17]

Answer:

T= 8.061N*m

Explanation:

The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:

T = Fr

where F is the force, r the radius.

if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.

Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:

r = $\sqrt{(0.38m)^2+(0.38m)^2}$

r = 0.5374m

Finally, replacing the value of r and F, we get that the maximun torque is:

T = 15N(0.5374m)

T= 8.061N*m

4 0

3 years ago

How to change V=72km/hr to m/s

lapo4ka [179]

Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).

Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.

(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)

= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)

= 20 meter/second

7 0

3 years ago

How can we calculate force

lutik1710 [3]

To calculate force, use the formula force equals mass times acceleration, or F = m × a. Make sure that the mass measurement you're using is in kilograms and the acceleration is in meters over seconds squared. When you've solved the equation, the force will be measured in Newtons.

8 0

2 years ago

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