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Leokris [45]
3 years ago
11

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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What is the H+ concentration for an aqueous solution with pOH = 4.01 at 25 ∘C? Express your answer to two significant figures an
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+

5 0
3 years ago
Does the excess reactant get used up completely in a reaction??
Alex787 [66]

Answer:

In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.

Explanation:

4 0
3 years ago
Arrange the compounds in order of decreasing magnitude of lattice energy:
Alik [6]

Answer:

The correct answer is CaO > LiBr > KI.

Explanation:

Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.  

With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.  

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.  

8 0
3 years ago
Sort the vocabulary word below. Make sure you have come to live session or watched the 5.03 recording. These words will be impor
Stells [14]

Answer:

Electric Current

This is a flow of electrons (such as the flow of electrons in a wire to light up a lamp).

Insulator

These stop the electricity from flowing (think rubber and glass).

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5 0
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

6 0
3 years ago
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