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Amanda [17]
3 years ago
10

A sample of ethane (C2H6)was combusted completely and the water that formed has a mass of 1.61 grams.How much ethane, in grams,

was in the sample?Put your answer in thespace provided belo socratic.org
Chemistry
1 answer:
ruslelena [56]3 years ago
5 0

Answer:

0.89 g of ethane

Explanation:

The balanced reaction equation is

C2H6(g) + 7/2 O2(g) -----------> 2CO2(g) + 3H2O(g)

From this balanced reaction

30g of ethane yields 54g of water

Therefore mass of ethane necessary to obtain 1.61g of water we have:

30 × 1.61/54 = 0.89 g of ethane

You might be interested in
Select the correct answer.
ivann1987 [24]

Answer:

C.

Explanation:

The electronic configuration of N (7 electrons): 1s² 2s² 2p³.

The orbital 1s is filled with two electrons and their spinning direction is opposite and also electrons of 2s.

3p contains (3 electrons) should fill the 3 orbitals firstly. Every orbital contains 1 electron and be in the same spin direction.

So, the right choice is c.

A is wrong because 2 electrons of 3p are paired in the first orbital before filling every orbital.

B is wrong because the 2 electrons of 1s and 2s are in the same direction and also 2 electrons of 3p are paired in the first orbital before filling every orbital.

D is also wrong the 2 electrons of 1s and 2s are in the same direction and the electron in the second orbital of 3p are in opposite direction of the other 2 electrons.

7 0
3 years ago
What is the molecular formula of a compound containing 89% cesium (Cs) and 11% oxygen (O) with a molar mass = 298 g/mol?
Maurinko [17]
M(Cs)=133 g/mol
M(O)=16 g/mol
M(CsxOy)=298 g/mol
w(Cs)=0.89
w(O)=0.11

CsxOy

x=M(CsxOy)w(Cs)/M(Cs)
x=298*0.89/133=2

y=M(CsxOy)w(O)/M(O)
y=298*0.11/16=2

Cs₂O₂  cesium peroxide
6 0
3 years ago
Read 2 more answers
Increase the pressure on the system
Molodets [167]

Answer:

What do you mean????????

6 0
3 years ago
How many atoms are found in the products of the following chemical equation?
coldgirl [10]

There are 13 atoms in the product

<h3>Further explanation</h3>

Given

Reaction

H2SO4 + 2KOH --> K2SO4 + 2H2O

Required

The number of atoms

Solution

In a chemical equation, there are reactants on the left and products on the right

Reactants : H2SO4 + 2KOH

Products : K2SO4 + 2H2O

The number of atoms is determined by their reaction coefficient and the subscript of the atoms in the compound

K2SO4 (coefficient = 1) :

K = 2 atoms

S = 1 atom

O = 4 atoms

Total atoms = 7 atoms

2H2O(coefficient = 2) :

H = 2 x 2 = 4 atoms

O = 2 x 1 = 2 atoms

Total atoms = 6 atoms

Total = 13 atoms

5 0
2 years ago
For the following reaction, 6.94 grams of water are mixed with excess sulfur dioxide . Assume that the percent yield of sulfurou
Alexxx [7]
<h3>Answer:</h3>

#a. Theoretical yield = 31.6 g

#b. Actual yield = 25.72 g

<h3>Explanation:</h3>

The equation for the reaction between sulfur dioxide and water to form sulfurous acid is given by the equation;

SO₂(g) + H₂O(l) → H₂SO₃(aq)

The percent yield of H₂SO₃ is 81.4%

Mass of water that reacted is 6.94 g

#a. To get the theoretical yield of H₂SO₃ we need to follow the following steps

Step 1: Calculate the moles of water

Molar mass of water = 18.02 g/mol

Mass of water = 6.94 g

But, moles = Mass/molar mass

Moles of water = 6.94 g ÷ 18.02 g/mol

                        = 0.385 mol

Step 2: Calculate moles of H₂SO₃

From the equation, the mole ratio of water to H₂SO₃ is 1 : 1

Therefore, moles of water = moles of H₂SO₃

Hence, moles of H₂SO₃ = 0.385 mol

Step 3: Theoretical mass of H₂SO₃

Mass = moles × Molar mass

Molar mass of H₂SO₃ = 82.08 g/mol

Number of moles of H₂SO₃ = 0.385 mol

Therefore;

Theoretical mass of H₂SO₃ = 0.385 mol ×  82.08 g/mol

                                             = 31.60 g

Thus, the theoretical yield of H₂SO₃ is 31.6 g

<h3>#b. Calculating the actual yield</h3>

We need to calculate the actual yield

Percent yield of H₂SO₃ is 81.4%

Theoretical yield is 31.60 g

But; Percent yield = (Actual yield/theoretical yield)×100

Therefore;

Actual yield = Percent yield × theoretical yield)÷ 100

                   = (81.4 % × 31.6) ÷ 100

                  = 25.72 g

The percent yield of H₂SO₃ is 25.72 g

6 0
3 years ago
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