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Citrus2011 [14]
3 years ago
10

FIND THE RATIO BY MASS OF THE COMBINING ELEMENTS IN THE FOLLOWING COMPOUNDS

Chemistry
1 answer:
boyakko [2]3 years ago
5 0
The given compound is Aluminum sulfate, Al2(SO4)3:

Molar masses:

Aluminum = 27 g/mol
Sulfur = 32 g/mol
Oxygen = 16 g/mol

The total molar mass is 342 g/mol
The ratio by mass of the elements:

Aluminum = 27*2/342 
                  = 0.16
Sulfur = (32*3)/342
           = 0.28
Oxygen = (16*12)/342
              = 0.56
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The particles begin to vibrate faster and more.

Explanation:

Adding heat to matter increases the energy, thus creating more movement. Eventually, the bucket will melt, turning to a liquid. While it is a sold, it still has particle movement, just not enough to break volume or shape.

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Different between a bird and a rat​
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How many Oxygen elements are in a carboxylic acid
Amanda [17]

Answer:

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Explanation:

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3 years ago
If the imaginary element Blobbium (Bb) formed an acid with the formula H₃Bb, what should this acid be named? *
sveticcg [70]

Answer:

C

Explanation:

The non metal in HCl is Chlorine and the name of the acid is Hydrochloric acid.

H = hydro

Cl = chloric

So I imagine this new acid, yet to be discovered even by the crew of the Star Ship enterprise, would be

HydroBlobbic acid

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4 0
3 years ago
The specific reaction rate of a reaction at 492k is 2.46 second inverse and at 528k 47.5 second inverse.find activation energy a
crimeas [40]

Answer:

Ea = 177x10³ J/mol

ko = 1.52x10^{19} J/mol

Explanation:

The specific reaction rate can be calculated by Arrhenius equation:

k = koxe^{-Ea/RT}

Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.

k depends on the temperature, so, we can divide the k of two different temperatures:

\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}

\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}

Applying natural logathim in both sides of the equations:

ln(k1/k2) = Ea/RT2 - Ea/RT1

ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)

R = 8.314 J/mol.K

ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)

ln(0.052) = (Ea/8.314)x(-1.38x10^{-4}

-1.67x10^{-5}xEa = -2.95

Ea = 177x10³ J/mol

To find ko, we just need to substitute Ea in one of the specific reaction rate equation:

k1 = koxe^{-Ea/RT1}

2.46 = koxe^{-177x10^3/8.314x492}

1.61x10^{-19}ko = 2.46

ko = 1.52x10^{19} J/mol

3 0
3 years ago
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