All categories

2 years ago

How many grams are in 6.53 moles of Mn?

Physics

1 answer:

Westkost [7]2 years ago

3 0

Answer:

359 g Mn

General Formulas and Concepts:

Dimensional Analysis
Reading the Periodic Table of Elements

Explanation:

Step 1: Define

6.53 mol Mn

Step 2: Find conversion

1 mol Mn = 54.94 g Mn

Step 3: Dimensional Analysis

$6.53 \hspace{3} mol \hspace{3} Mn(\frac{54.94 \hspace{3} g \hspace{3} Mn}{1 \hspace{3} mol \hspace{3} Mn} )$ = 358.758 g Mn

Step 4: Simplify

We are given 3 sig figs.

358.758 g Mn ≈ 359 g Mn

You might be interested in

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

Dmitry [639]

Answer:

Part a)

$F = 135.7 N$

Part b)

$F = 62.5 N$

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg + F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg + \mu(Fsin\theta)$

$F(cos\theta - \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta - \mu sin\theta}$

$F = \frac{55}{cos50 - 0.310(sin50)}$

$F = 135.7 N$

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg - F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg - \mu(Fsin\theta)$

$F(cos\theta + \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta + \mu sin\theta}$

$F = \frac{55}{cos50 + 0.310(sin50)}$

$F = 62.5 N$

6 0

3 years ago

Why might a balloon, that is inflated almost to its capacity, pop or explode on an extremely warm day?

REY [17]

On an extremely warm day, the balloon might pop because gases expand the hotter they get, and due to its temperature it is likely to pop if it is, indeed, nearly, if not completely, filled to its capacity. I hope this helps, have a nice day!

7 0

3 years ago

Read 2 more answers

It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

$Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\$

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

$\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\$

ω = 0.05 rad/s

7 0

3 years ago

What is the magnitude of the gravitational force of attraction to Jupiter exerts on IO

kotegsom [21]

The gravity force between Jupiter and Io will be 6.343 × 10²² N.

<h3>What is Newton's law of gravitation?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data;

Mass of Jupiter, $\rm m_j = 1.9 \times 10^{27} \ kg$

Mass of moon of Jupiter, $\rm m_{i_0}= 8.9 \times 10^{22} \ kg$ ]

The gravitational constant is, $\rm G = 6.67 \times 10^{-11 } \ m^3 kg^{-1} s^{-2}$

Distance between Jupiter and Io, R = 421,700 km = 4,217,00,000 m

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

The gravitational force is found as;

$\rm F = G \frac{ m_J m_{I_0}}{R^2} \\\\\ F = (6.67\times 10^{-11}) \frac{( (1.9\times 10^{27})\times (8.9\times 10^{22} )} { (421700000)^2}\\\\ F_g = 6.343 \times 10^{22} \ N$

Hence, the gravity force between Jupiter and Io will be 6.343 × 10²² N.

The complete question is

"Jupiter has a mass of 1.9 × 1027 kg, and its moon Io has a mass of 8.9 × 1022 kg. Their centers are separated by a distance of 421,700 km what is the force of gravity acting on Io? "

To learn more about Newton's law of gravitation, refer to the link.

brainly.com/question/9699135.

#SPJ1

8 0

2 years ago

Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni

Softa [21]

Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

$B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}$

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

$B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T$

hence, the magnitude of the magnetic field is 1.04μT

4 0

3 years ago