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CaHeK987 [17]
3 years ago
6

In an energy diagram, the reactants are at a higher potential energy compared to the products. Which of the following best descr

ibes this reaction?
It is endothermic and will have a positive enthalpy.
It is endothermic and will have a negative enthalpy.
It is exothermic and will have a positive enthalpy.
It is exothermic and will have a negative enthalpy.
It is endothermic and will have a neutral enthalpy.
Chemistry
2 answers:
Greeley [361]3 years ago
8 0

Answer:

The answer is D on Plato

Explanation:

vichka [17]3 years ago
4 0
Do you have a picture of where you go this question from?
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6.02 x 10 atoms of cu
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Answer:

what's the question

Explanation:

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3 years ago
What’s the difference between internal stimuli and external stimuli
oee [108]

Answer:

An external stimuli is a stimulus that comes from outside an organism and causes a reaction.An internal stimuli is a stimulus that comes from inside an organism.

3 0
3 years ago
Consider this reaction:
nadya68 [22]
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4 0
2 years ago
The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av
nlexa [21]

<u>Answer:</u> The percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{77}^{191}\textrm{Ir} isotope be 'x'. So, fractional abundance of _{77}^{193}\textrm{Ir} isotope will be '1 - x'

  • <u>For _{77}^{191}\textrm{Ir} isotope:</u>

Mass of _{77}^{191}\textrm{Ir} isotope = 190.9606 amu

Fractional abundance of _{77}^{191}\textrm{Ir} isotope = x

  • <u>For _{77}^{193}\textrm{Ir} isotope:</u>

Mass of _{77}^{193}\textrm{Ir} isotope = 192.9629 amu

Fractional abundance of _{77}^{193}\textrm{Ir} isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710

Percentage abundance of _{77}^{191}\textrm{Ir} isotope = 0.3710\times 100=37.10\%

Percentage abundance of _{77}^{193}\textrm{Ir} isotope = (1-0.3710)=0.6290\times 100=62.90\%

Hence, the percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

8 0
3 years ago
Given the reaction: HNO2 (aq) &lt;-&gt; H+ (aq) + NO2- (aq); write the ionization constant for the reaction.
Nuetrik [128]

Answer:

see explanation

Explanation:

HNO₂ ⇄ H⁺ + NO₂⁻

Ka = [H⁺][NO₂⁻]/[HNO₂]

4 0
3 years ago
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