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aliina [53]
3 years ago
12

Determine the molar mass of barium sulfide, BaS

Chemistry
1 answer:
matrenka [14]3 years ago
3 0

Answer: 169.39 g/mol

Explanation:

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The molar absorptivity for aqueous solutions of phenol at 211 nm is 6.17x103L/mol/cm. Calculate the linear range of phenol conce
Scilla [17]

Explanation:

The given data is as follows.

          \lambda = 211 nm,          \sum = 6.17 \times 10^{3} mol/L/cm

           l = 1 cm,             7% < Transmittance < 85%

Suppose the aqueous solution follows Lambert-Beer's law. Therefore,

                   Absorbance = -log \frac{\text{Percentage transmittance}}{100}

Hence, for 7% transmittance the value of absorbance will be as follows.

                  Absorbance = -log \frac{7}{100}

                           A_{1} = 1.155

For 85% transmittance the value of absorbance will be as follows.

                 Absorbance = -log \frac{85}{100}

                           A_{2} = 0.07058

According to Lambert-Beer's law.

                  A = \sum \times l \times C

where,       A = absorbance

                 \sum = molar extinction coefficient

                 C = concentration

Therefore, concentration for 7% absorbance is as follows.

                    A_{1} = \sum \times l \times C_{1}

                  C_{1} = \frac{1.155}{6.17 \times 10^{3} \times 1}

                                  = 0.187 \times 10^{-3} mol/L

                                  = 0.187 mmol/L

Concentration for 85% absorbance is as follows.

                   A_{2} = \sum \times l \times C_{2}

                  C_{2} = \frac{0.07058}{6.17 \times 10^{3} \times 1}

                                  = 0.01144 \times 10^{-3} mol/L

                                  = 0.01144 mmol/L

Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.

6 0
2 years ago
What is the name of the state when electrons absorb energy and move to a higher energy level?
Sav [38]
Excited state , as stated in quantum mechanics
6 0
3 years ago
Consider the tables below.
brilliants [131]

The right answer is

Table A Organic solvent

No Perfume No Fuel No Anesthetic No Adhesive Yes

6 0
3 years ago
Read 2 more answers
a 20.5g sample of cleaning detergent contains 8.61g of NH40H.CALCULATE the percentage composition of nitrogen in the cleaning de
AysviL [449]

Answer:

The mass percentage composition of nitrogen in the sample of the cleaning detergent is approximately 16.78%  

Explanation:

The given mass of the sample of the cleaning detergent, m₁ = 20.5 g

The mass of the ammonium hydroxide, NH₄OH in the detergent, m₂ = 8.61 g

The molar mass of NH₄OH = 35.04 g/mol

The molar mass of nitrogen, N = 14.01 g/mol

Therefore, the mass, m₃ of nitrogen, N, in 8.61 g of ammonium hydroxide, NH₄OH, is found as follows;

m₃ = (14.01/35.04) × 8.61 g = (402,087/118,800) g ≈ 3.44 g

The mass of nitrogen, N, in the ammonium hydroxide, NH₄OH, contained in the 20.5 g sample of the cleaning agent, m₃ ≈ 3.44 grams

The percentage composition of nitrogen in the sample of the cleaning detergent, %N is given as follows;

\% Composition = \dfrac{Mass \ of \ component}{Total \ mass \ of \ cleaning \ detergent} \times 100

Therefore;

%N ≈ ((3.44 g)/(20.5 g)) × 100 ≈ 16.78 %

The percentage composition of nitrogen, %N ≈ 16.78%.

6 0
2 years ago
A gas has a volume of 5.00 L at 0°C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to ea
Diano4ka-milaya [45]

Answer:

A = -213.09°C

B = 15014.85 °C

C = -268.37°C

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = 0°C  (273 K)

Final volume = 1100 mL, 280 L, 87.5 mL

Final temperature = ?

Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume =  87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

4 0
2 years ago
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