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Harman [31]
1 year ago
15

Is each of these statements true? If not, explain why.(m) For most reactions, ΔH°rnx is lowered by a catalyst.

Chemistry
1 answer:
barxatty [35]1 year ago
3 0

A  catalyst reduces H°rnx in most reactions. The answer is false

<h3>Do catalysts reduce delta H?</h3>

By reducing the activation energy required for the reaction to occur, a catalyst just modifies the route used to go from reactants to products. However, because it doesn't alter the state of the products or reactants, delta H is unaffected.

A catalyst reduces a reaction's activation energy, enabling a chemical reaction to occur. The number of reactant particles with energy above the activation energy increases as the temperature of a reaction rises.

learn more about catalyst refer

brainly.com/question/12507566

#SPJ4

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What part of an atom has no relative mass
Debora [2.8K]

Answer: Electrons

Explanation:

Most of an atoms' mass comes from the protons and neutrons that make up its nucleus. Electrons have the least mass of an atom's particles.

4 0
3 years ago
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You are determining the density of an unknown solid. You figure out the volume by water displacement. Then you weigh the solid,
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Answer:

The calculated density will be larger

Explanation:

The calculated density will be <u>larger</u>. Because, the volume is taken accurately, by the water displacement method. But, when we the took the mass, the water was present on the unknown solid. So, the mass of that water was added to the original mass of the solid. Hence, the mass measured was larger than the original mass. We, know from the formula of density that density is directly proportional to the mass of the object.

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Hence, the larger measured mass means the larger value of density.

8 0
2 years ago
When 18.5 g of HgO(s) is decomposed to form Hg(l) and O2(g), 7.75 kJ of heat is absorbed at standard-state conditions. What is t
taurus [48]

Answer:

The standard enthalpy of formation of HgO is -90.7 kJ/mol.

Explanation:

The reaction between Hg and oxygen is as follows.

\text{Hg(l)}+\frac{1}{2}{O_{2}\rightarrow \text{HgO(s)}

From the given,

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Amount of heat absorbed = 7.75 kJ

From the reaction,

The standard  enthalpy of formation = +7.75\times\frac{kJ}{18.5 g}\frac{216.59}{1mol} \,\,= +90.7 kJ/mol

During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.

For the formation of 1 mol of HgO , 90.7 kJ of energy is release

Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol

5 0
3 years ago
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3 0
3 years ago
Two major discoveries of the gold foil experiment
tiny-mole [99]
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