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WARRIOR [948]
4 years ago
11

a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter

Chemistry
1 answer:
ladessa [460]4 years ago
5 0

Answer:

The concentration of chloride ion is 2.82\times10^{-3}\;mol/L

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the Cl^- content 100 ppm suggests the presence of 100 mg of Cl^- in 1 L of solution.

The molar mass of Cl^- is equal to the molar mass of Cl atom as the mass of the excess electron in Cl^- is negligible as compared to the mass of Cl atom.

So, the molar mass of Cl^- is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of Cl^- present in 100 mg (0.100 g) of Cl^- is calculated as shown below:

N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol

So, there is 2.82\times10^{-3}\;mol of Cl^- present in 1 L of solution.

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eimsori [14]

Composer and performer of Old Occitan lyric poetryduring the High Middle Ages (1100–1350). Since the word troubadour is etymologically masculine, a female troubadour is usually called a trobairitz.


6 0
3 years ago
The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
A sample of potassium has an average atomic mass of 39.0983amu. There are three isotopic forms of potassium element in the sampl
igomit [66]

Answer:

Percent Composition of 41K = 6.7302%

Explanation:

The explination is in the image.

7 0
3 years ago
Mature elephants consume an average of 600 pounds of food per day. If an elephant
MAXImum [283]

Answer:

\large \boxed{\text{4 600 000 kg or 4600 Mg}}

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its  

Adult life = 55 - 10 + 1 = 46 yr

2. Convert years to days

\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}

3. Convert days to pounds of feed

\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}

4. Convert pounds to kilograms and megagrams

\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

5 0
3 years ago
When a saturated solution of NH4Br dissolved in 100 grams of water is cooled from 60°C to 30°C, how much NH4Br will precipitate?
stepladder [879]

Answer:

m_{precipitated}=24.8g

Explanation:

Hello,

In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:

m_{precipitated}=108g-83.2g\\\\m_{precipitated}=24.8g

Best regards.

8 0
3 years ago
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