About how long did it take for this population to reach the carrying capacity of this habitat

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

How many joules of heat are absorbed to raise the temperature of 435 grams of water at 1 atm from 25°C to its boiling point, 100

zepelin [54]

The energy absorbed may be calculated using:
Q = mcΔT
Where Q is the energy absorbed, m is the mass of water, c is the specific heat capacity of the water and ΔT is the change is the temperature of the water. Substituting the values,
Q = (435)(4.18)(100-25)
Q = 136,372 J

The energy absorbed is 136.4 kJ

7 0

3 years ago

When 1.82 moles of HCL reacts with excess MnO2, how many moles of Cl2 form

s2008m [1.1K]

The balanced reaction is:

MnO2(s) + 4HCl(aq) → Cl2(g) + MnCl2(aq) + 2H2O(l)

We are given the amount of hydrochloric acid to be used for the reaction. This will be the starting point for the calculations.

1.82 mol HCl ( 1 mol Cl2 / 4 mol HCl) = 0.46 mol Cl2

Therefore, 0.46 mol of chlorine gas is produced for the reaction of hydrochloric acid and manganese oxide.

5 0

3 years ago

Read 2 more answers

A molecule has two atoms that are bonded to the central atom and one lone pair of electrons around this central atom. What is th

alex41 [277]

The shape of the molecule is trigonal planar.
For example, H₂O:
..
O
/ \
H H

3 0

3 years ago

Read 2 more answers

Hello people ~ Help me with this question. Thanks in advance!

Alekssandra [29.7K]

Hello!