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Zolol [24]
3 years ago
8

PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEWSE HELP PLEASE H

ELP ANYBODY!!! CHEMISTRY DUE IN 5 MINUTES!! DIMENSIONAL ANALYSIS
How many moles of O2 are in the sample of 3.6 kilograms of air if 21% of the air is oxygen, O2, by mass?

WILL GIVE BRAINLIEST PLEASE ANYONE WHO KNOWS HOW TO DO MOLE CONVERSIONS AND DIMENSIONAL ANALYSIS

Chemistry
1 answer:
marishachu [46]3 years ago
7 0

Answer:

23.6 moles

Explanation:

From the question given above, the following data were obtained:

Mass of air = 3.6 Kg

Mass percentage of O₂ = 21%

Number of mole of O₂ =?

Next, we shall convert 3.6 Kg of air to grams (g). This can be obtained as follow:

1 kg = 1000 g

Therefore,

3.6 Kg = 3.6 Kg × 1000 / 1 kg

3.6 Kg = 3600 g

Next, we shall determine the mass of O₂ in the air. This can be obtained as follow:

Mass of air = 3600 g

Mass percentage of O₂ = 21%

Mass of O₂ =?

Mass of O₂ = 21% × 3600

Mass of O₂ = 21/100 × 3600

Mass of O₂ = 756 g

Finally, we shall determine the number of mole of O₂ in the sample of air. This can be obtained as follow:

Mass of O₂ = 756 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Number of mole of O₂ =?

Mole = mass /Molar mass

Number of mole of O₂ = 756 / 32

Number of mole of O₂ = 23.6 moles

Thus, the number of mole of O₂ in the

sample of air is 23.6 moles

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Can someone solve this for me I'm confused.
Artemon [7]

Answer:

310.53 g of Cu.

Explanation:

The balanced equation for the reaction is given below:

CuSO₄ + Zn —> ZnSO₄ + Cu

Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:

Molar mass of CuSO₄ = 63.5 + 32 + (16×4)

= 63.5 + 32 + 64

= 159.5 g/mol

Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g

Molar mass of Cu = 63.5 g/mol

Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g

Summary:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.

Thus, 310.53 g of Cu were obtained from the reaction.

6 0
3 years ago
What is the nuclear binding energy for thorium-234? 2.78 × 10-10 J 3.36 × 10-14 J 1.67 × 1017 J 5.35 x 10-23 J
grin007 [14]
The correct option is A.
To calculate the binding energy, you have to find the mass defect first.
Mass defect = [mass of proton and neutron] - Mass of the nucleus
The molar mass of thorium that we are given in the question is 234, the atomic number of thorium is 90, that means the number of neutrons in thorium is 
234 - 90 = 144.
The of proton in thourium is 90, same as the atomic number.
Mass defect = {[90 * 1.00728] +[144* 1.00867]} - 234
Note that each proton has a mass of 1.00728 amu and each neutron has the mass of 1.00867 amu.
Mass defect = [90.6552 + 145.24848] - 234 = 1.90368 amu.
Note that the unit of the mass is in amu, it has to be converted to kg
To calculate the mass in kg
Mass [kg] = 1.90368 * [1kg/6.02214 * 10^-26 = 3.161135 * 10^-27
To calculate the binding energy
E = MC^2
C = Speed of light constant = 2.9979245 *10^8 m/s2
E = [3.161135 * 10^-27] * [2.9979245 *10^8]^2
E = 2.84108682069 * 10^-10.
Note that we arrive at this answer because of the number of significant figures that we used.
So, from the option given, Option A is the nearest to the calculated value and is our answer for this problem.

8 0
3 years ago
Read 2 more answers
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
Identify each of the following sets of quantum numbers as allowed or not allowed in the hydrogen atom.
Westkost [7]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly recall the electron configuration of hydrogen:

1s^1

To realize that the principal quantum number is 1, the angular is 0 as well as the magnetic one; therefore we infer that all the given n's are not allowed, just l=0 is allowed as well as ml=0 yet the rest, are not allowed.

Best regards!

5 0
2 years ago
A airliner has an internal pressure of 1.00 ATM and temperatures of 25°C at take off if the temperature of the airlines routes t
Georgia [21]

Answer:

Explanation:

Using Gay lussac's law equation as follows;

P1/T1 = P2/T2

7 0
3 years ago
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