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Iteru [2.4K]
3 years ago
7

How many moles of aluminum oxide al2o3 are in a sample with a mass of 204.0

Chemistry
2 answers:
Leviafan [203]3 years ago
6 0

Answer:

2 moles

Explanation

Molecular mass of Al : 26.981539 u

Molecular mass of O: 15.999 u

The molecular mass of Al₂O₃ is:

= 101.960078 u

This can be approximated to 102 u.

1mole weighs 102 u

So, 2moles will weigh 2*102 = 204 u

erik [133]3 years ago
3 0

Answer:

2 moles

Explanation:

Let us first start by calculating the molecular mass of Al₂O₃.

The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.

Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.

Molecular mass of Al : 26.981539 u

Molecular mass of O: 15.999 u

Therefore, molecular mass of Al₂O₃ is:

= (2*26.981539) + (3*15.999) u

= 101.960078 u

This can be approximated to 102 u.

1mole weighs 102 u

So, 2moles will weigh 2*102 = 204 u

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What is the volume of water in 150ml of the 35% of sucrose with a specific gravity of 1.115?
anzhelika [568]

Answer:

The volume of the water is 108.71 mL

Explanation:

Step 1: Data given

Volume of water =150 mL = 0.150 L

concentration of sucrose solution 35 % w/w this means in 100 grams of water we have 35 grams of sucrose

specific gravity =1.115

Step 2: Calculate the density of the solution

Density = specific gravity * density of water

Density of solution = 1.115 * 1g/ mL

Density of solution = 1.115 g/ mL

Step 3: Calculate mass of the solution

Mass of solution = density ¨volume

Mass of solution = 1.115 g/ mL * 150 mL

Mass of solution = 167.25 grams

Step 4: Calculate mass of sucrose

35 % = 0.35 * 167.25 grams

Mass sucrose = 58.54 grams

Step 5: Calculate mass of water

Mass of water = mass of sample - mass of sucrose

Mass of water = 167.25 - 58.54 = 108.71 grams

Step 6: Calculate volume of water

Volume = mass / density

Volume = 108.71 grams / 1g/ mL

Volume = 108.71 mL = 0.10871 L

The volume of the water is 108.71 mL

5 0
3 years ago
Read 2 more answers
The molar heat of fusion of gold is 12.550 kJ mol–1. At its melting point, how much mass of melted gold must solidify to release
KATRIN_1 [288]

The mass of melted gold to release the energy would be  3, 688. 8 Kg

<h3>How to determine the mass</h3>

The formula for quantity of energy is given thus;

Q = n × HF

Where n represents number of moles

HF  represents  heat of fusion

To find the number of moles, we have

235.0 = n × 12.550

number of moles = \frac{235}{12. 550} = 18. 725 moles

Note that molar mass of Gold is 197g/ mol

Let's note that;

Number of moles = mass/ molar mass

Mass = number of moles × molar mass

Mass = 18. 725 × 197

Mass = 3, 688. 8 Kg

Thus, the mass of melted gold to release the energy would be  3, 688. 8 Kg

Learn more about molar heat of fusion here:

brainly.com/question/15634085

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8 0
1 year ago
Look at the following reaction. How could you increase the production of 2NH3(g)? N2(g) + 3H2(g) 2NH3(g) Increase the volume of
Zolol [24]
The correct option is B. To increase the production of ammonia, you have to increase the pressure of the system. Increase in pressure will result in increased production of ammonia because this will drive the chemical reaction forward.
5 0
3 years ago
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A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardm
VARVARA [1.3K]

<u>Answer:</u> The nuclear equations for the given process is written below.

<u>Explanation:</u>

The chemical equation for the bombardment of neutron to U-238 isotope follows:

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Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

The chemical equation for the first beta decay process of _{92}^{239}\textrm{U} follows:

_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta

The chemical equation for the second beta decay process of _{93}^{239}\textrm{Np} follows:

_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta

Hence, the nuclear equations for the given process is written above.

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Which statement is true about the patterns shown on the electromagnetic spectrum?
Rudiy27
A. As wavelength increases, radiant energy and frequency increase.
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