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konstantin123 [22]

3 years ago

15

The picture below shows a warm air mass caught between two cooler air masses. What is this type of front called? Plz, I really n

eed help.

1 answer:

pentagon [3]3 years ago

6 0

An occluded front forms when a warm air mass is caught between two cooler air masses.

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Which group on the periodic table contains elements in all 3 states of matter?

nordsb [41]

VII . A because it contains Br (l), Cl (g) and I (s)

4 0

3 years ago

Water containing large amounts of Mg2+and Ca2+ions is said to be hard because it is hard to make soap lather in the water. True

mafiozo [28]

Answer:True

Explanation:

Water is said to be hard when it contains calcium ions or magnesium ions dissolved in it. These ions are able to react with soap in such a way that the soap is prevented from forming lather with the water. Hard water occurs when water passes over calcium or magnesium bearing minerals and dissolves some of it. Hardness due to the presence of calcium ions can easily be removed by boiling the water.

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3 years ago

3mL of cyclohexanol (density = 0.9624 g/mL, Molecular weight = 100.158 g/mol) reacts with excess sulfuric acid to produce cycloh

TiliK225 [7]

Answer:

$n_{C_6H_{10}}=0.03molC_6H_{10}$

Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

$n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}$

Besides, the mass could be computed as well by using the molar mass of cyclohexene:

$m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}$

Even thought, the volume could be also computed by using its density:

$V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}$

Best regards.

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