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valentinak56 [21]
3 years ago
5

How many iron atoms are in the formula for iron(III) oxide?

Chemistry
1 answer:
velikii [3]3 years ago
8 0

Answer:

There is two iron atoms.

Explanation:

The formula iron oxide is Fe2O3 F e 2 O 3 therefore there would be two. Hope this helps. :)

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Lead ions and magnesium ions both have a charge of 2+. predict the ionic equation for this reaction.
FinnZ [79.3K]
Answer:
PbMg

Explanation:
Because they both have a charge of 2+, they can be reduced and cancel each other out because 2 and 2 can be reduced to 1
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Let us write the appropriate equilibria and associate the correction <img src="https://tex.z-dn.net/?f=K_b" id="TexFormula1" tit
rosijanka [135]

Explanation:

The relation between K_a\&K_b is given by :

K_w=K_a\times K_b

Where :

K_w=1\times 10^{-14} = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = K_{a1}=1.4\times 10^{-2}

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1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}

K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}

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The value of K_{b2}:

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3 0
3 years ago
Finding energy and wavelength?
deff fn [24]
The equation is
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——————————
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W= 4.87 x 10^-7 m

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3 0
3 years ago
DNA is packaged into these<br> compact units
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Explanation:

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4 0
3 years ago
Read 2 more answers
The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
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