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2 years ago

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For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou

ld NOT be the change in enthalpy (ΔH) and the activation energy (Ea), respectively, for the catalyzed reaction? Select all that apply.

1 answer:

Ronch [10]2 years ago

5 0

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

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A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol

qaws [65]

Answer:

Part a)

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

$d = \frac{1}{2}gt^2$

Part c)

$v_f = v_o - gt$

Part d)

$d = \frac{1}{2}gt^2$

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

$v_y = -gt$

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

$d = \frac{1}{2}gt^2$

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

$v_f = v_i + at$

$v_f = v_o - gt$

Part d)

distance from helicopter

$d = \frac{1}{2}gt^2$

8 0

2 years ago

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When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro

NISA [10]

Remember your kinematic equations for constant acceleration. One of the equations is $x_{f} = x_{i} + v_{i}(t) + \frac{1}{2} at^{2}$ , where $x_{f}$ = final position, $x_{i}$ = initial position, $v_{i}$ = initial velocity, t = time, and a = acceleration.

Your initial position is where you initially were before you braked. That means $x_{i}$ = 100m. You final position is where you ended up after t seconds passed, so $x_{f}$ = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was $v_{i}$ = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
$350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}$

Your acceleration is approximately $-7.2 \: m/s^{2}$ .

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2 years ago

Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

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1 year ago

13. An aircraft heads North at 320 km/h rel:

AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

$v' = v + v_a$

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

$v_a$ is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

$v_a = -80 km/h$ (since the air is moving from North)

Therefore, we find

$v'=+320 + (-80) = +240 km/h$ (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

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2 years ago

Anna needs to move a box of paperback books across the room. If she applies a force of 20 newtons to the box, what is the magnit

vlabodo [156]

Newton's third law tells us that for every force there is an equal and opposite force. This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.

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2 years ago

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