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IRISSAK [1]
3 years ago
14

How many moles of Na are in 11.9 grams of Na

Chemistry
2 answers:
hram777 [196]3 years ago
8 0

Answer:

sdgas

Explanation:

as dggsda

SVEN [57.7K]3 years ago
7 0
I think there are 0.5 moles of ana
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Which of the following is an acid?<br><br> Be(OH)2<br> HCl<br> LiBr<br> NH3
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Answer:

HCl is the correct answer

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Which element is oxidized in the reaction below? fe(co)5 (l) + 2hi (g) fe(co)4i2 (s) + co (g) + h2 (g)?
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Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂<span> (s) is x. For whole compound, the charge is zero.

Sum of oxidation numbers in all elements = Charge of the compound.

Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
 x + 4*0 + 2*(-1) = 0
 x + 0 - 2            = 0
                   x     = +2
Hence the oxidation state of Fe in product </span>Fe(CO)₄I₂ (s) is +2.

Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
   y        = 0

Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.
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What methods were used to clean up the Chernobyl and three mile island accidents?
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Electric evaporator has been used to clean up the Chernobyl and 3 mile island accident.

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The surface rocks on Venus are similar to ______ rocks on Earth.
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Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi
JulijaS [17]

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

<h3>How to determine the mass of Carbon </h3>
  • Mass of CO₂ = 1.47 g
  • Molar mass of CO₂ = 44 g/mol
  • Molar of C = 12 g/mol
  • Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

<h3>How to determine the mass of H</h3>
  • Mass of compound = 0.5 g
  • Mass of C = 0.4 g
  • Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

<h3>1. How to determine the empirical formula </h3>
  • C = 0.4 g
  • H = 0.1 g
  • Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

<h3>2. How to determine the molecular formula</h3>
  • Empirical formula = CH₃
  • Molar mass = 30 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Learn more about empirical formula:

brainly.com/question/24297883

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4 0
2 years ago
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