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In 1928, 47.5 g of a new element was isolated from 660 kg of the ore molybdenite. the percent by mass of this element in the ore

vitfil [10]

To take the percent by mass of this element, we use the formula:

% mass = (mass of element / mass of ore) * 100%

% mass = (47.5 g / (660 kg * 1000 g / kg)) * 100*

8 0

3 years ago

Read 2 more answers

If 2.0 mol of Zn is mixed with 3.0 mol<br> of HCl, which reactant will be limiting?

Soloha48 [4]

The following Balanced Reaction will take place:

Zn + 2HCl → ZnCl₂ + H₂

In the question, we have 2 moles of Zinc and 3 moles of HCl for this reaction

<u>Amount of HCl required to completely react with 2 moles of Zn:</u>

Since we need 2 moles of HCl for every mole of Zn, we will need 2(2) = 4 moles of HCl for every 2 moles of Zn

<u>Identifying the Limiting Reagent:</u>

But we are only given 3 moles of HCl where we need 4 moles to completely react.

So, since HCl is in less amount, it is the Limiting Reagent

8 0

3 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

3 years ago

300×175<br> Express your answer using the correct number of significant figures.

icang [17]

Answer:

300×175=52500

Explanation:

caculator

3 0

3 years ago

For the reaction below, initially the partial pressure of all 3 gases is 1.0atm. . 2NH3(g)--> N2(g) + 3H2(g) K, 0.83 1. When

erma4kov [3.2K]

Answer:

The reaction would shift toward the reactants

When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm

Explanation:

For the reaction:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Where K is defined as:

$K = \frac{P_{N_{2}}*P_{H_2}^3}{P_{NH_3}^2} = 0.83$

As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:

$Q = \frac{1atm*{1atm}^3}{1atm^2} = 1$

As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>

Thus, there are true:

<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>

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4 0

3 years ago