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3 years ago

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Use the mass spectrum of europium to determine the atomic mass of europium. where the peak representing eu-151 has an exact mass

of 150.91986 amu and percent intensity is assumed be to exactly 91.610%, and the peak representing eu-153 has an exact mass of 152.92138 amu and percent intensity of 100.00%. express your answer to two decimal places with the appropriate units.

2 answers:

umka21 [38]3 years ago

4 0

Answer:

The atomic mass of a Europium atom is 151.96445 amu.

From the given information:

Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.

Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.

Abundance of Eu-151 atom:

Abundance of Eu-153 atom:

Atomic mass of Europium atom:

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

Explanation:

slavikrds [6]3 years ago

3 0

Answer: The atomic mass of a Europium atom is 151.96445 amu.

From the given information:

Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.

Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.

Abundance of Eu-151 atom:

$X_{Eu-151}=\frac{0.9161}{0.9161+1.000}=0.4781$

Abundance of Eu-153 atom:

$X_{Eu-153}=\frac{1.000}{0.9161+1.000}=0.5219$

Atomic mass of Europium atom:

$A=(X_{Eu-151}\times150.91986+X_{Eu-153}\times152.92138)amu\\A=(0.4781\times150.91986+0.5219\times152.92138)amu=151.96445 amu$

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

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2 years ago

A ball of mass 2 kg is kept on the hill of height 3 km. Calculate the potential energy possessed by it ?

zysi [14]

We know that -

$P.E=m*g*h$

Where,

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$g = acceleration$ $due$ $to$ $gravity$

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3 km = 1000 * 3 meters = 3000 meters

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3 years ago

Read 2 more answers

A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person

vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 $\frac{m}{s}$

t=0.475s V=1.95 $\frac{m}{s}$

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

$sin(\alpha) =\frac{op}{h}$

$op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m$

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

$T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz$

c).

$T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz$

The period is the inverse of the time of the motion so, the T1 is faster that the T because

$t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s$

d).

The speed is the relation between the distance with time so:

$Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}$

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3 years ago

A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s^2 . She is driving at 20 m/s when su

QveST [7]

Answer:

The car stops after 32.58 m.

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 20 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6 m/s²

Time taken by the car to stop

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20}{-6}\\\Rightarrow t=3.33\ s$

Total Time taken by the car to stop is 0.5+3.33 = 3.83 s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 3.83+\frac{1}{2}\times -6\times 3.83^2\\\Rightarrow s=32.58\ m$

The car stops after 32.58 m.

Distance between car and obstacle is 50-32.58 = 17.42 m

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3 years ago