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olasank [31]
2 years ago
14

Calculate the mass of O2 produced after 20.0 grams of KClO3 have reacted?

Chemistry
1 answer:
IrinaVladis [17]2 years ago
5 0

Answer:

RFM \: of \: potassium \: chlorate = (39 + 35.5 + (16 \times 3)) = 122.5 \: g \\ 122.5 \: g \: of \: potassium \: chlorate \: produce \: (16 \times 3) \: g \: of \: oxygen \\ 20.0 \: g \: of \: potassium \: chlorate \: will \: produce \:  (\frac{(20.0 \times 16 \times 3)}{122.5} ) \: g \\  = 7.84 \: g \: of \: oxygen

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How many electrons in an atom can share the quantum numbers n = 1, m subscript l = 0? A. 1 B.2 C.3 D.6
Kisachek [45]
I believe it would be c: 3 because you can have the numbers -1,0,1.
6 0
3 years ago
Which of the following shows the valency of 1?
Oliga [24]

\huge \bf༆ Answer ༄

The element having valency of 1 is ~

  • Sodium (Na)
6 0
2 years ago
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.950 L of .420 M H2SO4 is mixed with .900 L of .260 M KOH. What concentration of sulfuric acid remains after neutralization?
kipiarov [429]
H₂SO₄:

V=0,95L
Cm=0,420mol/L

n = CmV = 0,42mol/L * 0,95L = 0,399mol

KOH:

V=0,9L
Cm=0,26mol/L

n = CmV = 0,26mol/L * 0,9L = 0,234mol

H₂SO₄            +           2KOH ⇒ K₂SO₄ + 2H₂O
1mol                :           2mol
0,399mol         :           0,234mol
                                    limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol

n = 0,282mol
V = 0,950L + 0,900L = 1,85L

Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
3 0
3 years ago
If a 100. -g sample of a hydrated compound contains 37.07-g sodium, 48.39-g carbonate and 14.54-g water, find the empirical form
Mumz [18]

he required empirical formula based on the data provided is Na2CO3.H2O.

<h3>What is empirical formula?</h3>

The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.

We have the following;

Mass of sodium = 37.07-g

Mass of carbonate = 48.39 g

Mass of water = 14.54-g

Number of moles of sodium = 37.07-g/23 g/mol = 2 moles

Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole

Number of moles of water = 14.54/18 g/mol = 1 mole

The mole ratio is 2 : 1: 1

Hence, the required empirical formula is Na2CO3.H2O

Learn more about empirical formula : brainly.com/question/11588623

3 0
2 years ago
Mercury has an atomic mass of 200.59 amu. calculate the mass of 3.0 x 10^10 atoms
Licemer1 [7]
To determine mass of the given number of atoms of mercury, we need a factor that would relate the number of atoms to number of moles. In this case, we use the Avogadro's number. It is a <span>number that represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. The number of units could be atoms, molecules, ions or electrons. To convert into mass, we use the given amu of mercury since it is equal to grams per mole. We calculate as follows:

</span>3.0 x 10^10 atoms ( 1 mol / 6.022 x 10^23 atoms ) ( 200.59 g / 1 mol ) = 9.99x10^-12 g Hg
5 0
3 years ago
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