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

Meme on newtons law of motion (should me made ur self not from any searh engine)

k0ka [10]

He will be a pilot and he will fly the plane over bridges fewwww

8 0

1 year ago

Provide an example of when momentum is conserved and explain your answer you can get 10 PTS if answered with a good explaination

dezoksy [38]

Answer:

$m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s$

Explanation:

Conservation of Momentum

The total momentum of a system of two particles is

$p=m_1v_1+m_2v_2$

Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

$p'=m_1v_1'+m_2v_2'$

The momentum is conserved if no external forces are acting on the system, thus

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

Let's put some numbers in the problem and say

$m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s$

$(8)(12)+(6)(4)=(8)(-6)+(6)(28)$

$96+24=-48+168$

120=120

It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s

4 0

2 years ago

If the only forces acting on a 2.0kg mass are F1 = (3i-8j)N and F2 = (5i+3j)N, what is the magnitude of the acceleration of the

belka [17]

Answer: 4.7m/s²

Explanation:

According to newton's first law,

Force = mass × acceleration

Since we are given more the one force, we will take the resultant of the two vectors.

Mass = 2.0kg

F1+F2 = (3i-8j)+(5i+3j)

Adding component wise, we have;

F1+F2 = 3i+5i-8j+3j

F1+F2 = 8i-5j

Resultant of the sum of the forces will be;

R² = (8i)²+(-5j)²

Since i.i = j.j = 1

R² = 8²+5²

R² = 64+25

R² = 89

R = √89

R = 9.4N

Our resultant force = 9.4N

Substituting in the formula

F = ma

9.4 = 2a

a = 9.4/2

a = 4.7m/s²

Therefore, magnitude of the acceleration of the particle is 4.7m/s²

3 0

2 years ago

You kick a soccer ball of mass 0.41 kg. the ball leaves your foot with an initial speed of 23 m/s. (a what is the magnitude of t

svetoff [14.1K]

Impulse describes the change of momentum. Since we don't know the momentum of the soccer ball before the hit, this question is hard to answer. If you assume the momentum of the ball before the hit was p = 0, then the change in momentum is just Δp = Impulse = mv.

7 0

2 years ago

Read 2 more answers

Explain when a falling object is in free fall.

Ksenya-84 [330]

There is no gravity

3 0

3 years ago

Read 2 more answers