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1 year ago

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Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi

s process?

2 answers:

morpeh [17]1 year ago

4 0

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = $\dfrac{0.029158}{4.03176}$ = 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

Xelga [282]1 year ago

4 0

Answer:

(a) 992.87 g

(b) $6.419\times 10^{14} J$

Solution:

As per the question:

Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g

(a) To calculate mass of He produced:

We know that:

Atomic mass of hydrogen is 1.00784 u

Also,

4 Hydrogen atoms constitutes 1 Helium atom

Mass of Helium formed after conversion:

$4\times 1.00784 = 4.03136 u$

Also, we know that:

Atomic mass of Helium is 4.002602 u

The loss of mass during conversion is:

4.03136 - 4.002602 = 0.028758 u

Now,

Fraction of lost mass, M' = $\frac{0.028758}{4.03136} = 0.007133 u$

Now,

For the loss of mass of 1000g = $0.007133\times 1000$ = 7.133 g

Mass of He produced in the process:

$M_{He} = 1000 - 7.133 = 992.87 g$

(b) To calculate the amount of energy released:

We use Eintein' relation of mass-enegy equivalence:

$E = M'c^{2}$

$E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J$

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A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the

timurjin [86]

Answer:

option C

Explanation:

given,

energy dissipated by the system to the surrounding = 12 J

Work done on the system = 28 J

change in internal energy of the system

Δ U = Q - W

system losses energy = - 12 J

work done = -28 J

Δ U = Q - W

Δ U = -12 -(-28)

Δ U = 16 J

hence, the correct answer is option C

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1 year ago

A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t

choli [55]

Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff (Ff -friction force)

Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25

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2 years ago

Read 2 more answers

A community calendar allows nonprofit organizations to add events and their intended purposes for community members to see.

lbvjy [14]

Out of all given choices, activities and donor list can be found on the given kind of informative website.

Answer: Option A and B

<u>Explanation:</u>

Non-profit need to do the examination and for sake activities that simply aren't successful. And afterwards, they have to look to a portion of these auxiliary changes that to discover increasingly productive approaches to make a feasible money related model for their social change work.

When occasions are crucial, allowed to visit, and concentrated on developing as well as managing present or potential significant donors (people, establishments, corporate pioneers, they can bode well. Yet, only in the event that you catch up with participants on a one-on-one premise to additionally put them in the association and in the long run request that they contribute or reestablish their commitments.

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1 year ago

A Body OF Volume 36cc Floats With 3/4 of its volume submerged in water . The density Of Body is

Radda [10]

Answer:

Density of body = 0.25g/cc

Explanation:

Given:

Volume submerged in water = 3/4

Find:

Density Of Body

Computation:

Density of body = fraction of body in liquid x density of water

Density of body = [1-3/4]1

Density of body = 0.25g/cc

8 0

1 year ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago