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Brums [2.3K]
3 years ago
13

Mercury has an atomic mass of 200.59 amu. calculate the mass of 3.0 x 10^10 atoms

Chemistry
1 answer:
Licemer1 [7]3 years ago
5 0
To determine mass of the given number of atoms of mercury, we need a factor that would relate the number of atoms to number of moles. In this case, we use the Avogadro's number. It is a <span>number that represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. The number of units could be atoms, molecules, ions or electrons. To convert into mass, we use the given amu of mercury since it is equal to grams per mole. We calculate as follows:

</span>3.0 x 10^10 atoms ( 1 mol / 6.022 x 10^23 atoms ) ( 200.59 g / 1 mol ) = 9.99x10^-12 g Hg
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During this reaction, Carbon Dioxide gas is evolved which causes the reaction mixture to fizz. The equation is given below.

NaHCO_{3}\ (aq) +  H_{3}C_{6}H_{5}O_{7} (aq)\rightarrow Na_{3}C_{6}H_{5}O_{7}(aq)+ CO_{2}(g)+H_{2}O (l)

Rate of the above reaction is affected by the Temperature.

As the temperature increases , the rate of the reaction increases. This happens because at higher temperature, the collisions between reacting species are more which result in formation of product in less time. This increases the rate of reaction.

We have been given equal volumes of water for each beaker. But the temperature of beaker c is 80°C which is the highest temperature. That means the reaction in beaker c is fastest.

Whereas beaker a is at lowest temperature (30°C) , therefore the reaction in beaker a would be slowest .

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3 years ago
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Answer:

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The nitrogen atoms in N2 participate in multiple bonding whereas those in hydrazine, N2H4, do not. a.) Draw Lewis structures for
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Answer:

a) Attached images.

b) N₂: sp; N₂H₄: sp³

c) N₂

Explanation:

a) In the attached images are the Lewis structures.

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<em>N₂H₄:</em> there is a single covalent bond between the N atoms.

b)

<em>N₂:</em> N has 2 electron domains. The corresponding hybridization is sp.  1 sp orbital form 1 sigma bonds whereas 2 p orbitals from 2 pi bonds.

<em>N₂H₄:</em> N has 4 electron domains. The corresponding hybridization is sp³. The 3 sp³ orbitals form 3 sigma bonds.

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