How are plastic containers and paper containers the same

Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

3 years ago

Using what you have learned about magnetic forces, potential and kinetic energy. Describe the changes

weeeeeb [17]

Answer:

reversible reaction or covalent bonds

7 0

2 years ago

Identify the environmental factor that caused the change in gene expression in the previous question. *

Arlecino [84]

It depends what Is the previous question

5 0

2 years ago

Read 2 more answers

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a flas

valentina_108 [34]

Answer:

Kp = 0.022

Explanation:

<em>Full question: ...With 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. </em>

<em />

The equilibrium of ammonia occurs as follows:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Where Kp is defined as:

$Kp = \frac{P_{N_2}P_{H_2}^3}{P_{NH_3}^2}$

<em>Where P represents partial pressure of each gas.</em>

<em />

As initial pressure of ammonia is 2.3atm, its equilibrium concentration will be:

P(NH₃) = 2.3atm - 2X

<em>Where X represents reaction coordinate</em>

<em />

Thus, pressure of hydrogen and nitrogen is:

P(N₂) = X

P(H₂) = 3X.

As partial pressure of hydrogen is 0.69atm:

3X = 0.69

X = 0.23atm:

P(NH₃) = 2.3atm - 2(0.23atm) = 1.84atm

P(N₂) = 0.23atm

P(H₂) = 0.69atm

$Kp = \frac{0.23atm*0.69atm^3}{1.84atm^2}$

8 0

3 years ago

a gas that exerts a pressure of 215 torr in a container with a volume of 51.0 mL will exert a pressure of ? torr when transferre

zhannawk [14.2K]

To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

P1V1 = P2V2,

Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:

(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr

The final pressure exerted by the gas would be 0.593 torr.

Hope this helps!

3 0

3 years ago