Ask question

Ask question

All categories

3 years ago

7

Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.75 L of a 0.130 M sucro

se solution is allowed to react for 190 minutes.

1 answer:

Advocard [28]3 years ago

3 0

Answer:125.84g

Explanation:Sucrose is dissacharides an organic compound in the class of carbonhydrate with the chemical formula C11H22O11.molar concentration is given by number of moles/Volume,this implies that moles=molar concentration ×Volume=0.130M×2.75L=0.3575moles.

Furthermore,number of moles=Mass of Sucrose/molecular Mass of Sucrose.

From it's formular C11H22O11, molecular Mass is the addition of the mass number which is 12 for C,2 for H and 16 for oxygen,O.so molecular Mass of Sucrose is (12×11)+(2×22)+(16×11)=352.

So mass =moles ×molecular mass=0.3575moles×352g/moles=125.84g

You might be interested in

Number of atoms in a 95.0 g sample of a gold nugget

crimeas [40]

Answer:

2.9 * 10^23 atoms Au

Explanation:

(95 g Au) * (1 mol Au/196.96657) * (6.02 * 10^23 atoms Au / 1 Mol Au)

2.9 * 10^23 atoms Au

5 0

3 years ago

A sample of argon gas at 55°C is under 845 mm Hg pressure. What will the new temperature be if the pressure is raised to 1050 mm

Maurinko [17]

Answer:

The final temperature at 1050 mmHg is 134.57 $^{\circ}C$ or 407.57 Kelvin.

Explanation:

Initial temperature = T = 55 $^{\circ}C$ = 328 K

Initial pressure = P = 845 mmHg

Assuming final to be temperature to be T' Kelvin

Final Pressure = P' = 1050 mmHg

The final temperature is obtained by following relation at constant volume

$\displaystyle \frac{P}{P'}=\displaystyle \frac{T}{T'} \\ \displaystyle \frac{845 \textrm{ mmHg}}{1050 \textrm{ mmHg}} = \displaystyle \frac{328 \textrm{ K}}{T'} \\T' = 407.57 \textrm{ Kelvin}$

The final temperature is 407.57 K

7 0

3 years ago

Which of following reactions shows the formation of a hydronium ion? H+ + H2O H3O+ H2O + CO2 H2CO3 NaOH Na+ + OH− H2O + NH3 OHâˆ

lesantik [10]

It can be formed by H+ ion and an H20 molecule also the chemical formula is H30+ or in other and easier way thus formula
H+ + H20=H30+

7 0

3 years ago

Read 2 more answers

What does the pH of a solution indicate

Nina [5.8K]

I think it’s B because I just know

8 0

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:

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

At inlet conditions, the mole fraction of chlorobenzene is:

$y_1 = \frac{731.44}{101.3}*\frac{101.3}{760}$

$y_1 = \frac{0.962 mol cholorobenzene}{mol saturated air}$

At outlet conditions , the mole fraction of chlorobenzene is:

$y_2 = \frac{60}{101.3}*\frac{101.3}{760}$

$y_2 = \frac{0.0789 mol cholorobenzene}{mol saturated air}$

Since there is no reaction, the total balance around the condensation is :

$n_1 = n_2 + n_3$

Let assume, that 100 moles of liquid chlorobenzene (CB) is condensed, therefore the equation becomes:

$n_1 = n_2$ + 100mol

The chlorobenzene balance using the mole fractions calculated above is :

$\frac{0.962mol CB}{mol(air)} * n_1 mol(air) = \frac{0.0789molCB}{mol(air)}*n_2mol(air) + 100CB$

substituting equation (1) into equation above; we have:

$\frac{0.962mol CB}{mol(air)} * n_1 mol(air) = \frac{0.0789molCB}{mol(air)}*(n_2-100)mol(air) + 100CB$

$0.962_n_1 mol = 0.0789_n_1mol + 100mol$

we can solve for n1, i.e ;

n1 = 104.3mol total air

Therefore the moles of chlorobenzene that will produce 100 moles of CB liquid is:

$n_C_B = 104.3 (moles) air * \frac{0.962 mol (CB)}{mol air}$

$n_C_B$ = 100.34mol CB

Now, calculating the percentage of chlorobenzene that condenses: we have;

% CB Condensation = $\frac{100mol}{100.34mol}*100%$

% CB Condensation = 99.7%

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

4 0

4 years ago