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zhuklara [117]
3 years ago
7

Calculate the number of vacancies in gold per cubic centimeter at 1000 K, where the density and atomic weight for gold at are 18

.63 g/cm3 and 196.9 g/mol, respectively, at 1000 K. The energy for vacancy formation in gold is 0.98 eV / atom. (Please express scientific notation using E to indicate the exponent - for example, 2.5E3
Chemistry
1 answer:
bazaltina [42]3 years ago
3 0

Answer:

autom is 18.63 g

Explanation:

there

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Are elements considered to be substances?
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What can natural selection accomplish? prove Darwin's idea of "survival of the fittest" change relative number of inhabitants wi
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Natural selection makes it so that an organism from a species has a better chance of surviving than an organism in the same species, but with different traits. Survival of the fittest proves that certain traits can help an organisms survive better than its peers. For example, say there was an allele that made bears have thicker fur than others. The year-round climate is extremely cold. The bears with thicker fur are more likely to survive due to the warmth their fur provides and will have more offspring. The bears with thinner fur will die more often and not produce healthy offspring due to the frigid temperature. Eventually, the allele for thin fur will die out because the bears with these alleles cannot survive in their habitat.
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3 years ago
Total number of electrons present in 1.8 grams of water is
Liono4ka [1.6K]

1 mole of water = 18 grams (you can find this by finding mass of two hydrogen and one oxygen which is (1*2) + 16 = 18)

1.8 grams = 0.1 moles

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(6.023*10^23)*(0.1)*10 = 6.023*10^23 electrons

5 0
3 years ago
Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.
sergejj [24]

The molecular formula of compound is Al(C_{2}H_{3}O_{2})_{3}.

Since, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6.023\times 10^{23} atoms of  Al(C_{2}H_{3}O_{2})_{3}.

Thus, according to molecular formula, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6\times 6.023\times 10^{23}=3.61\times 10^{24} atoms of oxygen atoms.

1 atom of oxygen will be present in \frac{1}{3.61\times 10^{24}} moles of Al(C_{2}H_{3}O_{2})_{3} . Thus,

2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}.

Molar mass of Al(C_{2}H_{3}O_{2})_{3} is 236 g/mol, mass can be calculated as follows:

m=n\times M=0.7285 mol\times 236 g/mol=172 g

Therefore, mass of Al(C_{2}H_{3}O_{2})_{3} will be 172 g.

5 0
3 years ago
Read 2 more answers
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