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3 years ago

The process of removal of oxygen in a chemical reaction is called?

Chemistry

2 answers:

solmaris [256]3 years ago

3 0

Answer:

Deoxygenation

Explanation:

This is when an oxygen atom is taken out of something, example when sand turns black in the mud flats. The sand is being deoxygenated by the ghost shrimp. Pulling oxygen from it.

Elza [17]3 years ago

3 0

Answer: Reduction

Explanation: because it's the removal of an O if it were an addition of an O is would be Oxidation.

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What's the boiling point of water at 0 atmospheric pressure?

nirvana33 [79]

Answer:

The water will evaporate and fly out of the bucket; the process will not stop until there is enough water vapor in the atmosphere that the vapor pressure stops the water from boiling further.

Explanation:

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2 years ago

Which area of chemistry best links the use of titanium and plastics in artificial bone and joint replacements?

kogti [31]

Answer:I think it’s material chemistry

Explanation:

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3 years ago

A mineral consisted of 29.4% calcium, 23.5% sulfur and 47.1% oxygen. What is the empirical formula?

inessss [21]

Step one calculate the moles of each element
that is moles= % composition/molar mass
molar mass of Ca = 40g/mol, S= 32 g/mol , O= 16 g/mol

moles of Ca = 29.4 /40g/mol=0.735 moles, S= 23.5/32 =0.734 moles, O= 47.1/16= 2.94 moles

calculate the mole ratio by dividing each mole with smallest mole that is 0.734
Ca= 0.735/0.734= 1, S= 0.734/0.734 =1, O = 2.94/ 0.734= 4
therefore the emipical formula = CaSO4

5 0

3 years ago

How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)

zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0

3 years ago

Under which conditions (in terms of kinetic energy) does boiling occur? How is the kinetic energy in the liquid increased?

Simora [160]

Explanation:

Boiling is defined as a process in which vapor pressure of a liquid substance becomes equal to the atmospheric pressure.

During this change liquid and vapors remain in equilibrium and the equation for this change is as follows.

$X(l) \rightleftharpoons X(g)$

Therefore, when boiling takes place then average kinetic energy of particles in liquid phase equals to the average kinetic energy of particles in vapor phase.

Hence, we can increase the kinetic energy of particles in liquid phase by increasing the temperature because kinetic energy is directly proportional to temperature as follows.

K.E = $\frac{3}{2}kT$

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3 years ago