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3 years ago

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RACTICE Uranium (molar mass = 283.03 g/mol) has the largest molar mass of any element naturally found on Earth. What is the mass

of 7.50 mol of uranium?

1 answer:

Marina86 [1]3 years ago

6 0

Answer:

<h3>The answer is 2122.73 g</h3>

Explanation:

The mass of a substance given it's molar mass and number of moles can be found by using the formula

mass = Molar mass × number of moles

From the question

molar mass = 283.03 g/mol

number of moles = 7.5 mol

So we have

mass = 283.03 × 7.5 = 2122.725

We have the final answer as

<h3>2122.73 g</h3>

Hope this helps you

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

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Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

3 years ago

The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve

solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d = 0.107 mm = 0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 / 4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0

3 years ago

Explain why the "X" chromosome expresses more traits than the "y" chromosome.

wel

Explanation:

The X chromosome (females have 2 of them while males have 1) is five times larger than the Y chromosome and has 10 times the number of genes.

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3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$

The frictional force on the block W₂, $F_{f2}$ ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ $F_{w2}$ = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + $\mathbf{F_{w2}}$

The frictional force from the ground, $\mathbf{F_{f1}}$ = N×μ + $\mathbf{F_{f2}}$ = P

Where;

P = The horizontal force applied to the block

P = (W₁ + $\mathbf{F_{w2}}$ ) × μ + $\mathbf{F_{f2}}$

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

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3 years ago

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air

fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

$u_y=0$ is the initial vertical velocity

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

$v_x = 10 m/s$

Therefore, the horizontal distance covered after a time t is

$d=v_x t$

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

$d=(10)(2.02)=20 m$

Learn more about projectile motion:

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3 years ago