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Volgvan
2 years ago
9

How do the percent compositions for C3H6 and C4H7 compare?

Chemistry
1 answer:
mariarad [96]2 years ago
3 0

A. They are the same

<h3>Further explanation</h3>

Given

C3H6 and C4H8

Required

The percent compositions

Solution

  • C₃H₆(MW = 42 g/mol)

%C = 3.12/42 x 100% = 85.71%

%H = 6.1/42 x 1005 = 14.29%

C₄H₈(MW=56 g/mol)

%C = 4.12/56 x 100% = 85.71%

%H = 8.1/56 x 100%=14.29%

So they are the same, because mol ratio of C and H in both compounds is the same, 1: 2

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O2- is the Lewis ______ in the following reaction
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O^{2-} is the Lewis base in the following reaction:

SO_3(g) + O^{2-}(aq)   →  SO_4^{2-}(aq)

<h3>What is lewis base? </h3>

Lewis base is a species that donates an electron pair.

In the above-given chemical equation, the O^{2-}is the lewis base which is donating an electron pair to the central atom of SO_3 to form SO_4^{-2}.

Hence, O^{2-} is the Lewis base in the following reaction:

SO_3(g) + O^{2-}(aq)   →  SO_4^{2-}(aq)

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3 0
1 year ago
Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:
Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: Al_{2}S_{3} +H_{2}O  _\to  Al(OH)_{3} + H_{2}S

First, balance the chemical equation

Al_{2}S_{3} + 6H_{2}O  \to 2Al(OH)_{3} + 3H_{2}S

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

n_{Al_{2}S_{3}  } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3}  } = \frac{158g}{150.16g/mol}   \\n_{Al_{2}S_{3} }=1.05 mol

From the chemical reaction , the ratio of molar is 3mol H_{2}S/1 mol Al_{2}S_{3}. So, the moles of hydrogen sulfide are:

n_{H_{2} O} =\frac{131g}{18.02g/mol}

       = 7.26mol

From the chemical reaction, the molar ratio is 3 mol H_{2}S/6 mol H_{2}O. So, the moles of hydrogen sulfide are:

Moles of H_{2}S formed = 7.26 mol H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }

Th liming reactant isAl_{2}S_{3} beacuse the mass of Al_{2}S_{3} forms less product than water. Therefore, the maximum number of moles of H_{2}S is 3.15 mol.  We know that molar mass of H_{2}S is 34.10g/mol. So, the maximum mass of H_{2}S formed is,

m_{H_{2}S } = n_{H_{2}S } \times Molar mass of H_{2}S

         = 3.15 mol \times 34.10g/mol

         = 107.4g

Now, multiplying the number of moles of Al_{2}S_{3} by the molar ratio between Al_2S_3 and H_2O which is 6mol H_2O/1mol Al_2S_3 we get the number of moles of H_2O reacted.

Moles of H_2O reacted = 1.05 mol Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}

                                     = 6.31 mol H_2O

The mass of H_2O is,

m_{H_{2} O} = 6.31 mol \times 18.02g/ mol

          = 114g

On subtracting, the mass of H_2O reacted from the given mass of H_2O is,

m_{H_2O} = (131-114)g

         = 17g

Hence, the excess remaining reactant is 17g

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2 years ago
What is the new pressure if the temperature outside is 296k and his weight causes the tire's volume to drop to 2.0 liters
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Question:

At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?

Answer:

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Explanation:

At standard temperature and pressure, we have:

P_1 = 1atm

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Determine the new pressure

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\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

This gives:

\frac{1 * 3.5}{273.15} =\frac{P_2*2.0}{296}

Solve for P_2

P_2 = \frac{296 * 1 * 3.5}{273.15*2.0}

P_2 = \frac{1036}{546.30}

P_2 \approx 1.896 atm

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Why are human matter?
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