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3 years ago

Balance the equations below by adding coefficients. Enter a number in each blank, entering "1" when necessary.

Chemistry

2 answers:

12345 [234]3 years ago

7 0

Answer:

1CO2 → 1C+ 1O2

1Fe2 + 1O2 → 2FeO

2Al + 3CuO → 1Al2O3 + 3Cu

Explanation:

1. CO2 → C+O2

On both sides we have 1x C and 2X O.

The equation is balanced

1CO2 → 1C+ 1O2

2. Fe2 + O2 → FeO

On the left side we 2x Fe, On the right side we have 1x Fe (in FeO). To balance the amount of Fe on both sides, we have to multiply FeO (on the right) by 2. If we multiply FeO by 2, we'll have 2x O on the right side. This is the same amount on the left side. The equation is balanced

1Fe2 + 1O2 → 2FeO

3. Al + CuO → Al2O3 + Cu

On the left side we have 1x Al, on the right side we have 2x Al. To balance the amount of Al we have to multiply Al (on the left) by 2.

We have 3x O on the right side, on the left side we have 1x O. To balance the amount of O on both sides, we have to multiply CuO (on the left) by 3.

We have 3x Cu on the left side (in 3CuO) and 1x Cu on the right side. To balance the amount of Cu we have to multiply Cu (on the right ) by 3.

Now the equation is balanced.

2Al + 3CuO → 1Al2O3 + 3Cu

neonofarm [45]3 years ago

3 0

Answer:

1. CO₂ → C + O₂

2. Fe₂ + O₂ → 2FeO

3. 2Al + 3CuO→ Al₂O₃ + 3Cu

Explanation:

1. 1 mol of CO₂ decomposes to 1 mol of C and 1 mol of oxygem

2. 1 mol of Fe₂ reacts with 1 mol of oxygen to produce 2 moles of iron (II) oxide

3. 2 moles of Al, reacts with 3 moles of cupper(II) oxide to produce 1 mol of aluminum dioxide and 3 moles of cupper

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3 years ago

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at 58.3°C, the

Elan Coil [88]

Answer:

The slope is 4661.4K

(B), the intercept is 18.156

The vapor pressure of chlorobenzene is 731.4mmHg

The percentage of chlorobenzene originally in the vapor that condenses is = 99.7%

Explanation:

Two sets of conditions (a and b) are observed for L1 and L2 at two different temperatures.

$T_{a}$ = 58.3°C

$L1_{a}$ = 747mmHg

$L2_{a}$ = 52mmHg

$T_{b}$ = 110°C

$L1_{a}$ = 577mmHg

$L2_{b}$ = 222mmHg

We need to convert the temperatures from Celsius to Kelvin (K)

$T_{a}$ = 58.3°C + 273.2

$T_{a}$ = 331.5k

$T_{b}$ = 110°C + 273.2

$T_{b}$ = 383.2k

We then calculate the vapor pressures of the chlorobenzene at each set of conditions by measuring the difference in the mercury levels.

$P^{0}$ = $P_{atm} - (P_1 -P_2)$

The vapor pressure under the first set of conditions is:

$P^{a}$ = 755mmHg - (747mmHg - 52mmHg)

$P^{a}$ = 60mmHg

The vapor pressure under the second set of conditions is:

$P^{b}$ = 755mmHg - (577mmHg = 222mmHg)

$P^{b}$ = 400mmHg

First question say we should find ΔH(slope) and B(intercept) in the Clausius- Clapeyron equation:

Using the Formula :

$In_p^{0} = \frac{- (delta) H}{RT} + B$

where (delta) H = ΔH

The slope of the $In_p^{0} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

Calculating the slope; we have:

$\frac{- (delta) H}{R} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

$\frac{- (delta) H}{R}$ = $\frac{331.5 * 383.2 * In (400/60)}{(331.5-383.2)}$

$\frac{- (delta) H}{R}$ = -4661.4k

$\frac{ (delta) H}{R}$ = 4661.4k

The intercept can be derived from the Clausius-Clayperon Equation by making B the subject of the formula. To calculate the intercept using the first set of condition above; we have:

B = $In_p_{1} + \frac{ (delta) H}{RT}$

B = $In 60 + \frac{4661.4}{331.5}$

B = 18.156

Thus the Clausius-Clayperon equation for chlorobenzene can be expressed as:

In P = $\frac{-4661.4k}{T} + 18.156$

In question (b), the air saturated with chlorobenzeneis 130°C, converting he temperature of 130°C to absolute units of kelvin(k) we have;

T = 130°C + 273.2

T = 403.2k

Calculating the vapor pressure using Clausius-Clapyeron equation: we have;

$In_p_{0}$ = $In_p_{0} = \frac{-4661.4}{403.2} + 18.156$

$In_p_{0}$ = 6.595

$p_{0}$ = $e^{6.595}$

$p_{0}$ = 731.mmHg

The vapor pressure of chlorobenzene is 731.4mmHg

The diagram of the flowchart with the seperate vapor and liquid stream can be found in the attached document below.

Afterwards, both the inlet and outlet conditions contain saturated liquid.

From the flowchart, the vapor pressure of chlorobenzene at the inlet and outlet temperatures are known:

$P_1(130^{0}C)$ = 731.44mmHg

$P_1(58.3^{0}C)$ = 60mmHg

To calculate the percentage of the chlorobenzene originally in the vapor pressure that condenses; we must first calculate the mole fractions of chlorobenzene for the vapor inlet and outlet using Raoult's Law:

$y_1 = \frac{P^0 (T)}{P}$