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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

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A 1.5-mole sample of an ideal gas is gently cooled at constant temperature 320 K. It contracts

Alona [7]

Answer:

V₂ = 14.07 L

Explanation:

As this gas is cooled at constant temperature of 320 K, this means that we are on an isothermal process, and according to the 1st law of thermodynamics:

Q = W (1)

And as the temperature is constant, we can use the following expression to calculate the Work done:

W = nRT ln(V₁/V₂) (2)

However, as Q = W, we can replace heat into the above expression and then solve for V₂:

Q = nRT ln(V₁/V₂)

Replacing we have:

1200 = (1.5 * 8.314 * 320) ln(19/V₂)

1200 = 39907.2 ln(19/V₂)

ln(19/V₂) = 1200/3990.72

ln(19/V₂) = 0.3007

19/V₂ = e^(0.3007)

V₂ = 19 / e^(0.3007)

Hope this helps

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3 years ago

Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend

Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

$\rho_{se}$ = Density of seawater = 1030 kg/m³

$\rho_{f}$ = Density of freshwater = 1000 kg/m³

$\rho_{sh}$ = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

$F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N$

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

$F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N$

The lift force required in a river is 34.7123 N

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3 years ago

At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

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3 years ago

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Potassium-A
Bromine-B
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7 0

3 years ago