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3 years ago

19. Which of the following chemical equations represents

Chemistry

1 answer:

Lesechka [4]3 years ago

6 0

Answer:

C₃H₈ (g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (l)

(option D. with the proviso that the subscripts of propane's chemical formula must be corrected)

Explanation:

Propane is the saturated hydrocarbon, alkane, with chemical formula C₃H₈ or CH₃CH₂CH₃.

The complete combustion of the hydrocarbons yield carbon dioxide (CO₂) and water (H₂O).

The chemical equation that represents this combustion is:

C₃H₈ (g) + O₂(g) → CO₂ (g) + H₂O (l) (skeleton equation: unbalanced)

Once you balance it, you get:

C₃H₈ (g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (l)

Left side Right side

C 3 3

H 8 4×2 = 8

O 5×2 = 10 3×2 + 4 = 10

That equation corresponds to the option D. of the list, with the proviso that the subscripts of propane's chemical formula must be corrected

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Freezing a candy bar is an example of a physical change. True or false

Debora [2.8K]

Answer: This is true.

Explanation:

It is true because if becomes frozen, then it is physically harder to melt...

6 0

2 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

Answer: The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

Explanation:

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i

alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

Explanation:

The pH of the solution = 2.46

$pH=-\log[H^+]$

$2.46=-\log[H^+]$

$[H^+]=0.003467 M$

$HA\rightleftharpoons H^++A^-$

Initially

0.0144 0 0

At equilibrium

(0.0144-x) x x

The expression if an dissociation constant is given by :

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(0.0144-x)}$

$x=[H^+]=0.003467 M$

$K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}$

$K_a=1.099\times 10^{-3}$

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

3 0

3 years ago

What is 11/6 + 11/6 in fractions

Andreyy89

3 2/3 is your answer

6 0

3 years ago

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