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soldi70 [24.7K]
2 years ago
14

Which will get spoiled faster: plum in water or plum in the open?

Chemistry
1 answer:
lara31 [8.8K]2 years ago
3 0

Answer:

It depends

(plum will spoil more quickly at warm temperatures)

Explanation:

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Exactly 1 mol Hg2(NO3)2 contains how many moles of Hg , N , and O ?
marin [14]

In exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Since the molecular formula of Hg₂(NO₃)₂ shows that for every mole of the substance, we have 2 moles of Hg, 2 moles of N and 6 moles of O.

So, in exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Learn more about number of moles here:

brainly.com/question/3935424

6 0
2 years ago
When the [H+] in a solution is 1.7 × 10−9 M, what is the pOH ? 8.80 5.23 14 8.60
kumpel [21]
PH = -log[H+]
pH = -log[1,7×10^-9]
pH = 8,77

pH + pOH = 14
pOH = 14 - 8,77
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5 0
3 years ago
Read 2 more answers
What do measurements of the mass of the Milky Way indicate?
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8 0
3 years ago
At standard pressure, the difference between the freezing point and the boiling point of water, in Kelvin degrees, is D. 373 A.
Sergio [31]
The answer to this question is A. 100
 

7 0
3 years ago
what is the order of decreasing bond length for a C-C bond composed of the following molecular orbitals?I. sp^3 - sp^3II. sp^2 -
skelet666 [1.2K]

Answer:

The decreasing order of bond length in the carbon - carbon bonds will be:

C-C>C=C>C\equiv C

Explanation:

Bond length is defined as average distance between two nuclei of  bonded atoms in a molecule.Bond length is inversely  proportional to the number of bonds present between two atoms.

\text{Bond length} \propto \frac{1}{\text{Number of bonds}}...[1]

Bond energy is defied as amount of energy required to break apart the bond of 1 mole of molecule into their individual atom. Bond energy is directly proportional to the number of bonds present between two atoms.

\text{Bond Energy} \propto \text{Number of bonds}..[2]

From [1] and [2]:

\text{Bond length} \propto \frac{1}{\text{Bond energy}}

C-C: sp_3 hybridized

C=C: sp_2 hybridized

C\equiv C: sp hybridized

Extent of overlapping of orbitals in these hybridization;

C\equiv C> C=C > C-C

Higher the overlapping of orbital more closer will be both atoms to each other and shorter will be the bond lenght.

So, the decreasing order of bond length in the carbon - carbon bonds will be:

C-C>C=C>C\equiv C

8 0
2 years ago
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