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Alexeev081 [22]
3 years ago
15

A microspectrophotometer measures which of the following?

Chemistry
1 answer:
iragen [17]3 years ago
4 0

Answer : The correct option is, Light absorption

Explanation :

Microspectrophotometer : It is a type of instrument that measures the spectra of the microscopic sample by using the different wavelengths of an electromagnetic radiation.

It is designed to measure the the spectrum of the microscopic sample. It is also measure the transmittance, absorbance, polarization and fluorescence of the microscopic sample.

From the given options, the option light absorption is the correct option.

Hence, a microspectrophotometer measures the light absorption.

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When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant =  

m= molality = \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579

8.30^0C=1\times K_f\times 0.579

K_f=14.3^0C/m

Let Mass of solute (KBr) = x g

8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}

x=58.2g

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

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Ammonia (NH3) reacts with oxygen to form nitric oxide (NO) and water vapor: 4NH3 + 502 4NO + 6H2O b) When 20.0 g NH3 and 50.0 g
Solnce55 [7]

Answer: a) . Ammonia is the limiting reagent

b. Oxygen is left over and 0.1375 g of oxygen is left over.

c. The theoretical yield of NO is 35.29 g.

d. The theoretical yield of H_2O is 31.74 g.

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

For NH_3

Given mass of ammonia = 20.0 g

Molar mass of ammonia = 17.031 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{20.0g}{17.031g/mol}=1.17mol

For O_2

Given mass of oxygen gas = 50.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{50.0g}{32g/mol}=1.6mol

The chemical equation for the reaction is

4NH_3+5O_2\rightarrow 4NO+6H_2O

By Stoichiometry of the reaction:

4 moles of ammonia reacts with = 5 moles of oxygen

So 1.17 moles of ammonia will react with = \frac{5}{4}\times 1.17=1.4625mol of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.

Thus ammonia is considered as a limiting reagent because it limits the formation of product.

1. By Stoichiometry of the reaction:

4 moles of ammonia produces = 4 moles of NO

1.17 moles of ammonia will produce = \frac{4}{4}\times 1.17=1.17moles of NO

Mass of NO=moles\times {\text{Molar Mass}}=1.17\times 30=35.29g

Thus Theoretical yield of NO is 35.29 grams.

2. By Stoichiometry of the reaction:

4 moles of ammonia produces = 6 moles of H_2O

1.2 moles of ammonia will produce = \frac{6}{4}\times 1.2=1.8moles of H_2O

Mass of H_2O=moles\times {\text{Molar Mass}}=1.8\times 18.015=31.74g  H_2O

Thus Theoretical yield of H_2O is 31.74 grams.

8 0
3 years ago
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