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3 years ago

HRISTINA HERRERA: Attempt 1

Chemistry

1 answer:

Lunna [17]3 years ago

8 0

Answer:

7.5 g of hydrogen gas reacts with 50.0 g oxygen gas to form 57.5 g of water.

Explanation:

Here we have the check if the mass of the reactants is equal to the mass of the products.

Reactants

$7.5+50=57.5\ \text{g}$

Products

$57.5\ \text{g}$

The data is consistent with the law of conservation of matter.

Reactants

$50+243=293\ \text{g}$

Products

$206+97=303\ \text{g}$

The data is not consistent with the law of conservation of matter.

Reactant

$17.7+34.7=52.4\ \text{g}$

Products

$62.4\ \text{g}$

The data is not consistent with the law of conservation of matter.

Only the first data is consistent with the law of conservation of matter.

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EleoNora [17]

I believe pH = -log[H+]

Also, 14 = pH + pOH

Therefore pH = 14 - pOH

pH = 14 - 4.01

pH = 9.99

9.99 = -log[H+]

Solve for H+

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3 years ago

Does the excess reactant get used up completely in a reaction??

Alex787 [66]

Answer:

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Explanation:

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Alik [6]

Answer:

The correct answer is CaO > LiBr > KI.

Explanation:

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The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.

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3 years ago

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Stells [14]

Answer:

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3 years ago

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

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3 years ago