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yanalaym [24]
3 years ago
7

1. Air boiled out of water as steam is rich in ?

Chemistry
1 answer:
Colt1911 [192]3 years ago
4 0

Answer:

<u><em>1. The air boiled out of water as steam is rich in water vapor, </em></u><em><u>H₂O</u></em><u><em>(g)</em></u>

<u><em></em></u>

<u><em>2. Benzene reacts with hydrogen in the presence of nickel catalyst at 180°C to give cyclohexane.</em></u>

Explanation:

<em></em>

1. When water boils liquid water passes to gaseous phase.

Water in gaseous phase is named water vapor or steam.

This happens around 100ºC, depending on the atmospheric pressure.

The air that is around the boiling water will carry that steam, becomind an air rich in water vapor, H₂O(g)

2. Benzene has chemical formula C₆H₆.

It is a cyclic unsaturated hydrocarbon. Benzene contains 3 double bonds.

When unsaturated hydrocarbons are heated in the presence of nickel catalyst, the double bonds are broken, forming single bonds and adding the needed hydrogen to the chain: one to each carbon that had a double bond transformed into single bond.

Thus, the product of the catalytic hydrogenation of benzene, C₆H₆,  is cyclohexane, C₆H₁₂

The image attached, taken fron the internet, can help you.

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Answer:

a. Kenya's growth seems to be a large population influx, an increase in the pre-reproductive stage.

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A tank with volume of 2.4 cu ft is filled with Methane to a pressure of 1500 psia at 104 degrees F. Determine the molecular weig
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Explanation:

It is known that equation for ideal gas is as follows.

               PV = nRT

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     Pressure, P = 1500 psia,     Temperature, T = 104^{o}F = 104 + 460 = 564 R

     Volume, V = 2.4 cubic ft,      R = 10.73 psia ft^{3}/lb mol R

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

                n = \frac{mass}{\text{molar mass}}

            m = n \times W

                = 0.594 \times 16.04

                = 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

  • We will calculate the density as follows.

                d = \frac{PM}{RT}

                    = \frac{1500 \times 16.04}{10.73 \times 564}

                    = 3.975 lb/ft^{3}

  • Now, calculate the specific gravity of the gas as follows.

  Specific gravity relative to air = \frac{\text{density of methane}}{\text{density of air}}

                         = \frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}

                         = 51.96

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How many pili are in a bacterial cell
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