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3 years ago

8

Combustion analysis of toluene, a common organic solvent, gives 6.45 mg of CO2 and 1.51 mg of H2O. If the compound contains only

carbon and hydrogen, what is its empirical formula?

1 answer:

yan [13]3 years ago

6 0

Answer:

Empirical formula - $C_{7}H_{8}$

Explanation:

From the given,

Mass of $CO_{2}$ = 6.45 mg= 0.00645 g

Molar mass of carbon = 12 g/mol

Molar mass of $CO_{2}$ = 44 g/mol

Given mass of $H_{2}O$ = 1.51 mg= 0.00151 g

Molar mass of water = 18 g/mol

Molar mass of hydrogen = 1.0 g/mol

$0.00645\times \frac{1molCO_{2}}{44gCO_{2}}\times \frac{1mol\,C}{1mol\,CO_{2}}= 1$

$0.00151\times \frac{1molH_{2}O}{44gH_{2}O}\times \frac{2mol\,H}{1mol\,H_{2}O}= 1.4$

1:1.4

Those two values are multiplied by two.(molar ratio -7)

$1:1.4\times 7= 7:8$

Therefore, empirical formula - $C_{7}H_{8}$

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pychu [463]

Answer:

The slope of the line is $-\frac{10}{7}$ .

Explanation:

The slope of the line ( $m$ ) is the change in dependent variable ( $y$ ) divided by the change in independent variable ( $x$ ):

$m = \frac{y_{f}-y_{o}}{x_{f}-x_{o}}$ (1)

If we know that $(x_{o}, y_{o}) = (0, 2)$ and $(x_{f}, y_{f}) = (1.4, 0)$ , then the slope of the line is:

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2 years ago

The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.

exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. $Na_{2} CO_{3}$

Major Species at B:

1. $Na_{2} CO_{3}$

2. $NaHCO_{3}$

Major Species at C:

1. $NaHCO_{3}$

Major Species at D:

1. $NaHCO_{3}$

2. $H_{2}CO_{3}$

Major Species at E:

1. $H_{2}CO_{3}$

Major Species at F:

1. $H_{2}CO_{3}$

b) pH calculation:

At Halfway point B:

pH = pK $a_{1}$ + log[ $CO_{3}$ $.^{-2}$ ]/[H $CO_{3}$ $.^{-1}$ ]

pH = pK $a_{1}$ = 6.35

Similarly, at halfway point D.

At point D,

pH = pK $a_{2}$ + log [H $CO_{3}$ $.^{-1}$ ]/[H2 $CO_{3}$ ]

pH = pK $a_{2}$ = 10.33

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2 years ago

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cestrela7 [59]

Answer:

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2 years ago

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timama [110]

H2S donates a proton, therefore it is a Brønsted-Lowry base; CH3NH2 accepts a proton, so it’s a Brønsted-Lowry base.

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3 years ago

The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 17

yawa3891 [41]

Answer:

a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

b. 0.957 g

Explanation:

Step 1: Write the balanced equation

2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

Step 2: Convert 130.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15

K = 130.0°C + 273.15

K = 403.2 K

Step 3: Calculate the moles of O₂

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K

n = 2.21 × 10⁻³ mol

Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂

The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.

Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO

The molar mass of HgO is 216.59 g/mol.

4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g

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2 years ago