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Phantasy [73]
3 years ago
12

A calibration curve relating absorbance(y) to concentration (x, measured in mol/L) of y= 0.36x-0.011 Use the calibration curve t

o determination of the concentration of a solution with an absorbance of 0.02.
Chemistry
1 answer:
katrin [286]3 years ago
3 0

Answer:

The concentration is 0.025 M.

Explanation:

We will use beer law to find the concentration.

A = εmCl

This law relate to the linear equation Y = mx + b in following way,

A = y = absorbance

m = εm = slope or extinction coefficient in beer law.

x = c = concentration

and l is length which is 1 cm

Thus equation can be written as,

A = εmc + b

A - b / εm = c

in order to find the concentration we will abstract the intercept from absorbance and then divide it by the slop.

0.02 - 0.011 / 0.36 = c

c =0.009

c = 0.025 M.

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A sample of 0.281 gg of an unknown monoprotic acid was dissolved in 25.0 mLmL of water and titrated with 0.0950 M NaOH NaOH. The
tensa zangetsu [6.8K]

Answer:

98.6 g/mol.

Explanation:

Equation of the reaction

HX + NaOH--> NaX + H2O

Number of moles = molar concentration × volume

= 0.095 × 0.03

= 0.00285 moles

By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.

Molar mass = mass ÷ number of moles

= 0.281 ÷ 0.00285

= 98.6 g/mol.

5 0
3 years ago
2 HCI → H2 + Cl2<br> Endothermic or exothermic
Oksana_A [137]

Answer:

Exothermic

Explanation:

I think it is exothermic because heat is being released

6 0
3 years ago
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of
Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH}  =0.1900gO

4. Compute the moles of oxygen by using its molar mass:

n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1

6. Search for the closest whole number (in this case multiply by 2):

C_3H_6O_2

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

M=12*3+1*6+16*2=74g/mol

Which is about three times in the molecular formula, for that reason, the actual formula is:

C_9H_{18}O_6

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0
3 years ago
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