1 scientific law and 1 scientific theory ?

The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process

denis23 [38]

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

7 0

3 years ago

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

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

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago

If 60. liters of hydrogen gas at 546 K is cooled to 273 K at constant pressure, the new volume of the gas would be

Korvikt [17]

Answer:30 L

Explanation:

Initial Volume

=

V

1

=

60

l

i

t

e

r

Initial Temperature

=

T

1

=

546

K

Final Temperature

=

T

2

=

273

K

Final Vloume

=

V

2

=

?

Sol:-

Since the pressure is constant and the question is asking about temperature and volume, i.e,

V

1

T

1

=

V

2

T

2

⇒

V

2

=

V

1

⋅

T

2

T

1

=

60

⋅

273

546

=

60

2

=

30

l

i

t

e

r

⇒

V

2

=

30

l

i

t

e

r

Hence the new volume of the gas is

30

l

i

t

e

r

6 0

3 years ago

Hydrates that have a low vapor pressure and remove moisture from air are said to be ___. Question 8 options: effloresce hygrosco

-Dominant- [34]

Answer:

Hygroscopic

Explanation:

An hygroscopic substance is one that absorbs moisture from the atmosphere and becomes wet. Their ability to remove water from air is less than that of deliquescent substances. Most of the solid hygroscopic substances forms pasty substances and not solutions like the deliquescent compounds.

Examples are sodium trioxonitrate(v), copper(ii) oxide e.t.c

Efflorescence compounds gives off their water of crystallization to the atmosphere.

4 0

2 years ago

A compound composed of 3. 3 % h, 19. 3 % c, and 77. 4 % o has a molar mass of approximately 60 g/mol. What is the molecular form

Brut [27]

The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as Carbonic acid.