Choose the false statement about an object that has mass.

A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat

Jet001 [13]

Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.

6 0

4 years ago

The vaporization of 1 mole of liquid water (system) at 100.9 C, 1.00 atm, is endothermic.

yan [13]

Answer:

a) The work done is 10.0777 kJ

b) The water's change in internal energy is -122.1973 kJ

Explanation:

Given data:

1 mol of liquid water

T₁ = temperature = 100.9°C

P = pressure = 1 atm

Endothermic reaction

T₂ = temperature = 100°C

1 mol of water vapor

VL = volume of liquid water = 18.8 mL = 0.0188 L

VG = volume of water vapor = 30.62 L

3.25 moles of liquid water vaporizes

Q = heat added to the system = -40.7 kJ

Questions: a) Calculate the work done on or by the system, W = ?

b) Calculate the water's change in internal energy, ΔU = ?

Heat for 3.25 moles:

$Q_{1} =3.25*(-40.7)=-132.275kJ$

The work done:

$W=-nP*delta(V)=-3.25*101.33*(0.0188-30.62)=10077.6637J=10.0777kJ$

The change in internal energy:

$delt(U)=W+Q_{1} =10.0777-132.275=-122.1973kJ$

7 0

3 years ago

Which of the following best describes the location of element Tin?

Tomtit [17]

Answer:

The location of element tin is

Group 14, Period 5

Explanation:

7 0

3 years ago

Read 2 more answers

A solution is created by dissolving 11.0 grams of ammonium chloride in enough water to make 235 mL of solution. How many moles o

Amanda [17]

Answer:

The correct answer is 0.206 moles

Explanation:

According to the given scenario, the calculation of the number of moles of ammonium chloride is available in the resulting solution is given below:

Given that

Amount of $NH_4Cl$ is 11.0 grams

And, the volume is 235 mL

Now the molar mass of $NH_4Cl$ is 53.49g/mol

So, the number of moles presented is

= 11.0 ÷ 53.49

= 0.206 moles

hence, the number of moles of ammonium chloride are available in the resulting solution is 0.206 moles

7 0

3 years ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

4 years ago