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Ksenya-84 [330]
3 years ago
13

< Question 3 of 25

Chemistry
1 answer:
Nonamiya [84]3 years ago
5 0
The answer is 360ms I hope this helps
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each element is designated by its __________ which is usually from the first letters of the elements name
klemol [59]

Answer:

each element is designated by its <u>Chemical Symbol </u>which is usually from the first letters of the elements name

Explanation:

7 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
The North Star is the last star in the Ursa Minor constellation.<br> True<br> False
Alja [10]
The answer is true. It is the last star.
4 0
3 years ago
What is an atomic weight?
S_A_V [24]
The average weight of an atom of an element, formerly based on the weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the weight of an oxygen atom, but after 1961 based on 1/12 the weight of the carbon-12 atom.
3 0
3 years ago
How much energy is required to raise the temperature of 10.7 grams of gaseous helium from 22.1 °C to 39.4 °C ?
Rainbow [258]

Answer:

Q = 2640.96 J

Explanation:

Given data:

Mass of He gas = 10.7 g

Initial temperature = 22.1°C

Final temperature = 39.4°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 39.4°C - 22.1°C

ΔT = 17.3°C

Q = 10.7 g× 14.267 J/g.°C ×  17.3°C

Q = 2640.96 J

5 0
3 years ago
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