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3 years ago

What period of history worsened air and water pollution in the 1800s?

Chemistry

1 answer:

notka56 [123]3 years ago

7 0

Answer:

The Industrial Revolution

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What is the capacity to do work known as?

yarga [219]

Answer:

it is energy

Explanation:

Energy is the ability to do work

6 0

3 years ago

Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002

marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

$pH=-\log[H^+]$

$=-\log[0.0001901 M]=3.72$

3 0

3 years ago

What is the area of the throat that contains the vocal cords and produces

aalyn [17]

C larynx hope this helps

7 0

4 years ago

Assuming your average walking speed is 1.4 m/s, what time should you leave your house to arrive at school (0.84 km away) at 7:15

Juli2301 [7.4K]

Answer: 6:50

Explanation:

I think that is correct is use the math correctly u could get 6:50 like i got or 7:00.

3 0

3 years ago

Read 2 more answers

Eading and

4vir4ik [10]

Answer: 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles$

The balanced chemical reaction is:

$2AuCl_3\rightarrow 2Au+3Cl_2$

According to stoichiometry :

2 moles of $AuCl_3$ produce = 3 moles of $Cl_2$

Thus 0.242 moles of will produce= $\frac{3}{2}\times 0.242=0.363mol$ of $Cl_2$

Mass of $Cl_2$ = $moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g$

Thus 25.8 g of $Cl_2$ will be produced from the decomposition of 73.4 g of $AuCl_3$

5 0

3 years ago