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alexgriva [62]
3 years ago
12

Would you describe acetone in the gaseous state as a vapor or gas

Chemistry
1 answer:
murzikaleks [220]3 years ago
3 0

Acetone is a vapor, since it is normally a liquid at room temperature, which is the definition of a vapor.

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List one reason as to why historical growth has slowed in recent years.
hodyreva [135]
Lower fertility and longer lifespans steadily increased the potential labor force relative to the total population
4 0
3 years ago
The electron pair in a C-F bond could be considered Question 3 options: closer to C because carbon has a larger radius and thus
Leokris [45]

Answer:

closer to F because fluorine has a higher electronegativity than carbon

Explanation:

Electronegativity refers to the ability of an atom in a bonding situation to draw the shared electrons of the bond closer to itself.

Electronegativity increases across the period and decreases down the group. A highly electronegative atom draws the shared electron pair of a bond towards itself.

When two atoms are bonded together, the electron pair is always drawn closer to the atom that has a higher electronegativity.

Hence, the electron pair in a C-F bond could be considered closer to F because fluorine has a higher electronegativity than carbon.

4 0
2 years ago
Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what
Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0
3 years ago
An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.
Stels [109]

Answer: The percent yield of the reaction is 77.0 %

Explanation:

2Pb+O_2\rightarrow 2PbO

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles

\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles

According to stoichiometry:

2 moles of Pb produces = 2 moles of PbO_2

2.18 moles of Pb is produced by=\frac{2}{2}\times 2.18=2.18moles of PbO_2

Mass of PbO_2 =moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6

percent yield =\frac{374.7g}{486.6g}\times 100=77.0\%

3 0
3 years ago
You read a primary source and a secondary source that discuss the same
shutvik [7]

Answer:

i think it's D tbh, just cus it was the scientist who did the work

4 0
2 years ago
Read 2 more answers
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