Why is it necessary to have different types of equipment for the use in the laboratory?

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H

Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}$

where,

$\Delta T_f$ = change in freezing point

$k_f$ = freezing point constant (for benzene} = $5.12^0Ckg/mol$

m = molality

Putting in the values we get:

$0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}$

${\text{ Mass of solute in g}}=0.028g$

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.

4 0

3 years ago

Stock naming please help

andre [41]

Answer:

Can you tell me the question in the comments on this answer or like how you do it then ill answer you in the comments under this answer

8 0

2 years ago

The amount of oxygen bound to hemoglobin _____.

Luba_88 [7]

The amount of oxygen bound to hemoglobin is 98.5%

7 0

3 years ago

Read 2 more answers

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$

(b) Since, the rate equation is as follows.

Rate = $K [OCl^{-}][l^{-}]$

Let us assume that ( $[OCl^{-}] = [l^{-}]$ )

Putting the given values into the above equation as follows.

$1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2$

$1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})$

K = $\frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}$

= $60.4 M^{-1}sec^{-1}$

Hence, the value of rate constant for the given reaction is $60.4 M^{-1}sec^{-1}$ .

(c) Now, we will calculate the rate as follows.

Rate = $K [OCl^{-}][l^{-}]$

= $60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})$

= $6.52 \times 10^{5}$

Therefore, rate when $[OCl^{-}] = 1.8 \times 10^{3}$ M and $[I^{-}]= 6.0 \times 10^{4}$ M is $6.52 \times 10^{5}$ .

8 0

3 years ago

Explain reaction of metals with the following with the help of chemical equations. i) with acids ii) with oxygen iii) with water

Amiraneli [1.4K]

i. When an acid reacts with metal, a salt and hydrogen are produced:

ii. When oxygen and metal react, metal oxide forms also known as rust

iii. metal and water produce hydrogen gas

6 0

3 years ago