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3 years ago

How to balance barium and oxygen yield barium oxide

1 answer:

4 0

Barium : $Ba^{+2}$ with +2 being the charge
Oxygen : $O^{-2}$ with -2 being the charge

The given equation can be written as:
Ba + O = BaO

Since the sum charges of Barium and Oxygen equals 0, there is no need to add subscripts.

Both Ba and O appear on the left and right side of the equation once, so there is no need to add a coefficient.

Ba + O = BaO is balanced

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What is the resulting formula unit when aluminum (Al) and phosphorus (P) bond?

Delvig [45]

<span>aluminium phosphide AlP</span>

7 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

If the mass of a material is 40 grams and the volume of the material is 9 cm^3, what would the density of the material be?

solniwko [45]

Mass = m = 40 grams
Volume = V = 9 cm³
Density = d = ?

Density is defined as the ratio of mass and volume.

So,

d = m/V

Using the values, we get

d = 40/9 = 4.44 g/cm³

This means the density of material would be 4.44 g/cm³

3 0

3 years ago

Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be

vlabodo [156]

Answer:

The theoretical yield of alum is 19.873 grams and percent yield for this alum synthesis is 45.33%.

Explanation:

Mass of bottle = 10.221 g

Mass of bottle with aluminium pieces = 11.353 g

Mass of aluminium = 11.353 g - 10.221 g = 1.132 g

Mass of alum and bottle = 19.230 g

Mass of alum = 19.230 g - 10.221 g = 9.009 g

Experimental yield of alum = 9.009 g

Theoretical yield of alum:

$2Al(s) +2 KOH(aq) +4H_2SO_4(aq)+10 H_2O(l) \rightarrow 2 KAl(SO_4)_2.12 H_2O(s)+3H_2(g)$

Moles of aluminium = $\frac{1.132 g}{27 g/mol}=0.041926 mol$

According to reaction, 2 moles of aluminum gives 2 moles of alum.

Then 0.041926 mol aluminium will give :

$\frac{2}{2}\times 0.041926 mol=0.041926 mol$ of alum.

Mass of 0.041926 moles of alum:

0.041926 mol × 474 g/mol= 19.873 g

Theoretical yield of alum = 19.873 g

Percentage yield:

$\% Yield =\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the alum:

$\% Yield =\frac{ 9.009 g}{19.873 g}\times 100=45.33\%$

The theoretical yield of alum is 19.873 grams and percent yield for this alum synthesis is 45.33%.

7 0

3 years ago

In order to make 0.500 L of a 0.475 M solution, you will need to weigh out how many grams of sucrose (MM = 342 g/mol)?

elixir [45]

The first step to solve this problem is to multiply the volume of solution times its concentration to find the number of moles needed, remember that M=mol/L:

$0.500L\cdot\frac{0.475mol}{L}=0.2375mol$

Now, use the molar mass of sucrose to find the number of grams needed to make the solution. This is, multiply the number of moles needed times the molar mass:

$0.2375mol\cdot\frac{342g}{mol}=81.225g$

It means that to make 0.500L of a 0.475M you will have to weigh 81.225g of sucrose.

4 0

1 year ago