Some species of moose are adapted to eat grasses and flowers in the summer and pine needles in the winter.

A sample of nitric acid has a mass of 8.2g. It is dissolved in 1L of water. A 25mL aliquot of this acid is titrated with NaOH. T

AveGali [126]

Answer:

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 ----- 1 mole

8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution ---- 0.13014 moles of HNO3

25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH ----- 1000 mL Solution

0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.

4 0

2 years ago

Hardness in groundwater is due to the presence of metal ions, primarily Mg2 and Ca2 . Hardness is generally reported as ppm CaCO

Maslowich

Answer:

0.0144 M

1440 ppm

Explanation:

Ca(In)²⁺ + EDTA → Ca(EDTA)²⁺ + In

We use the volume of EDTA consumed in the titration to calculate the moles of Ca⁺² ions:

0.012 L * 0.0600 M * $\frac{1molCa^{+2}}{1molEDTA}$ = 7.20x10⁻⁴ mol Ca⁺²

Now we calculate the molarity:

7.20x10⁻⁴ mol Ca⁺² / 0.050 L = 0.0144 M

To calculate in ppm, we use the moles of Ca⁺² and convert to mg of CaCO₃:

7.20x10⁻⁴ mol Ca⁺² = 7.20x10⁻⁴ mol CaCO₃

7.20x10⁻⁴ mol CaCO₃ * 100g/mol * $\frac{1000mg}{1g}$ = 72 mg CaCO₃

Finally, the concentration in ppm is:

72 mg CaCO₃ / 0.050L = 1440 ppm

5 0

2 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser: