Copper is a heavy metal that can harm many people. This process of heavy metals building in your body is called bioaccumulation and the process of this building in the food chain is called biomagnification. Its hard to rid of this heavy metal because it is very dense, therefore, cannot breakdown in our bodies. Heavy metals get consumed by producers and then reach to the top of the pyramid (us) so we have the highest risk of getting diseases and so on. Make sure you explain that heavy metals get passed from industrial purposes where companies do not dispose of waste properly.
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4 0

2 years ago

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A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the

tino4ka555 [31]

Answer:

2.1 × 10⁻¹ M

2.0 × 10⁻¹ m

Explanation:

Molarity

The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:

3.9 g × (1 mol/93.13 g) = 0.042 mol

The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:

M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M

Molality

The moles of solute are 0.042 mol.

The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:

200 mL × 1.05 g/mL = 210 g = 0.210 kg

The molality of aniline is:

m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m

5 0

2 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}