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Leya [2.2K]
3 years ago
9

When you have a different isotope of the same element, what is changed

Chemistry
1 answer:
seropon [69]3 years ago
6 0

Answer:

mass

Explanation:

An isotope is an element with different numbers of neutrons.  Changing the number of neutrons won't affect the height, volume, or periodic table alignment.  It will only affect the mass.

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Which of the following is not a conjugate acid-base pair?
-Dominant- [34]

Answer:

A. H3O+/OH−

Explanation:

A conjugate acid-base pair are a pair of molecules that differ in 1 H⁺

A. The conjugate pair of H₃O⁺ = H₂O not OH⁻

B. The conjugate pair of NH₄⁺ is NH₃

C. The conjugate pair of C₂H₃O₂⁻ is HC₂H₃O₂

D. The conjugate pair of H₂SO₃ is HSO₃⁻

That means right option, that is not a conjugate acid-base pair, is:

<h3>A. H3O+/OH− </h3>

5 0
3 years ago
Can someone help pls
Anastasy [175]
It is either concave lens or mirror but since this is reflecting out rather than reflecting back it has to be concave lens. hope that helps! :’)
3 0
4 years ago
Consider the reaction: 2BrF3(g) --&gt; Br2(g) + 3F2(g)
riadik2000 [5.3K]

Answer : The entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)

The expression used for entropy change of reaction (\Delta S^o) is:

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]

where,

\Delta S^o = entropy change of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

\Delta S_f^0_{(Br_2)} = 245.463 J/mol.K

\Delta S_f^0_{(F_2)} = 202.78 J/mol.K

\Delta S_f^0_{(BrF_3)} = 292.53 J/mol.K

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]

\Delta S^o=268.74J/K

Now we have to calculate the entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of BrF_3 has entropy change = 268.74 J/K

So, 1.62 moles of BrF_3 has entropy change = \frac{1.62}{2}\times 268.74=217.68J/K

Therefore, the entropy change of reaction for 1.62 moles of BrF_3 reacts at standard condition is 217.68 J/K

3 0
4 years ago
How is a kilogram of cotton balls similar to a kilogram of bricks
Nutka1998 [239]

Answer:

because they are both a kilogram

Explanation:

7 0
3 years ago
Which conclusion could be made from Ernest Rutherford’s gold foil experiment ?
mote1985 [20]
From the Rutherford's gold foil experiment one can conclude that nucleus was very small in size, as compared to the atoms. In the experiment Rutherford discovered that, the atom contains a very small nucleus where all of its positive charge of the atom is present.
3 0
4 years ago
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