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3 years ago

The gas usually filled in the electric bulb is

Chemistry

2 answers:

Mariulka [41]3 years ago

8 0

Answer:

1st answer: A. nitrogen

2nd answer: A. Sodium carbonate

Explanation:

Alex17521 [72]3 years ago

7 0

The gas usually filled in the electric bulb is

OPTION A

Washing soda is the common name for

OPTION A

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galben [10]

Answer: Area of a Triangle Equals Base x Height / 3

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The absorbance (????)(A) of a solution is defined as ????=log10(????0????) A=log10⁡(I0I) where ????0I0 is the incident‑light int

bulgar [2K]

Answer:

The absorbance of the myoglobin solution across a 1 cm path is 0.84.

Explanation:

Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times c\times l$

where,

A = absorbance of solution

c = concentration of solution

$\epsilon$ = Molar absorption coefficient

l = path length

$I_o$ = incident light

$I$ = transmitted light

Given :

l = 1 cm, c = 1 mg/mL , $\epsilon = 15,000 M^{-1}cm^{-1}$

Molar mass of myoglobin = 17.8 kDa = 17.8 kg/mol=17800 g/mol

(1 Da = 1 g/mol)

c = 1 mg /mL = ${1mg /mL}{\text{Molar mass of myoglobin}}$

$c = \frac{1 mg/mL}{ 17800 g/mol} = 5.6179\times 10^{-5} mol/L$

1 mg = 0.001 g, 1 mL = 0.001 L

$A= 15,000 M^{-1}cm^{-1}\times 5.6179\times 10^{-5} mol/L\times 1 cm$

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The absorbance of the myoglobin solution across a 1 cm path is 0.84.

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3 years ago

In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul

BARSIC [14]

Answer: The mass of water that should be added in 203.07 grams

Explanation:

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

m = molality of barium iodide solution = 0.175 m

$m_{solute}$ = Given mass of solute (barium iodide) = 13.9 g

$M_{solute}$ = Molar mass of solute (barium iodide) = 391.14 g/mol

$W_{solvent}$ = Mass of solvent (water) = ? g

Putting values in above equation, we get:

$0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g$

Hence, the mass of water that should be added in 203.07 grams

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3 years ago

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