Why soft drink is a mixture and water is compound

A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach with gas chromatography u

devlian [24]

Answer:

0.0136mg DDT / g spinach

Explanation:

Quantification in chromatography by internal standard has as formula:

RF = Aanalyte×Cstd / Astd×Canalyte (1)

Where RF is response factor, A is area and C is concentration

Replacing with first experiment values:

RF = 5019×3.20mg/L / 8179×6.37mg/L

RF = 0.308

In the next experiment, final concentration of chloroform was:

11.45mg/L × (1.25mL / 25.00mL) = 0.5725mg/L

From (1), it is possible to write:

Aanalyte×Cstd / Astd×RF = Canalyte

Replacing:

6821×0.5725mg/L / 14061×0.308 = Canalyte

Canalyte = 0.9017mg/L

as the sample was made from 0.750mL of extract. Concentration of extract is:

0.9017mg/L × (25.00mL / 0.750mL) = 30.06mg/L. As the extract has a volume of 2.40mL:

30.06mg/L × 2.40x10⁻³L = 0.07213mg of DDT in the extract

As the extract was made from 5.29g of spinach:

0.07213mg of DDT in the extract / 5.29g spinach = 0.0136mg DDT / g spinach

5 0

4 years ago

If an obect looks blue it reflects____ waves

amid [387]

All but blue cuh ♿️

6 0

3 years ago

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

Step 1: Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

Step 2: Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

Step 3: Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

Step 4: Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0

4 years ago

Calculate the freezing temperature of the following solution of 0.50 M glucose (a covalent compound). Assume that the molality o

kirza4 [7]

Answer:

-0.93 °C

Explanation:

Hello,

The freezing-point depression is given by:

$T_f-T_f^*=-iK_{solvent}m_{solute}$

Whereas $T_f$ is the freezing temperature of the solution, $T_f^*$ is the freezing temperature of the pure solvent (0 °C since it is water), $i$ the Van't Hoff factor (1 since the solute is covalent), $K_{f,solvent}$ the solvent's freezing point depression point constant (in this case $1.86 C\frac{kg}{mol}$ ) and $m_{solute}$ the molality of the glucose.