what mass of sulfur dioxide would be produced if 64 tonnes of sulfur was completely reacted with oxygen gas?

sladkih [1.3K]

Answer:

The smallest tiny particles of matter which can't be divided further is called atom, i.e., an atom is the smallest building block of matter. For example: Sodium (Na), Hydrogen (H), Oxygen (O), etc.

7 0

3 years ago

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Researchers are studying a fish population in a small pond that has algae as the only producer in the ecosystem. The scientists

Pie

Explanation:

Algae is a group of a photosynthetic eukaryotic organisms that plays a very important role on the ecosystem. It forms the lower base of the food chain and supports all the life forms of the ecosystem. It takes up water and carbon dioxide from the atmosphere to prepare food during the process of photosynthesis and maintains a balance in the ecosystem.

In the context, if the algae disappears, the fish might die as the small fishes feeds on this algae. This is a concern as the algae is main component of food source and forms the base of the food chain. If there will be no algae there will be no life form on earth, as all species depend on each other to survive.

In order to prevent this from occurring, we must protect our algae. We must plant more trees, keep our water bodies healthy, we should not litter our ponds and lakes. We should also help these algae to grow and survive so that other life forms can feed them and thus maintain the ecosystem.

7 0

3 years ago

6. A chemist needs 457.8 g of KMnO4 to make a solution. How many moles of KMnO4 is that?

Sedaia [141]

Answer:

6) 2.9 moles of KMnO4

7) 1.36×10^-3 moles

8) 2.8×10^-5 moles

Explanation:

6) Number of moles is given as mass/molar mass

Mass of KMnO4= 457.8

Molar mass of KMnO4= 158.034 g/mol

Number of moles of KMnO4= 457.8 g/ 158.034 g/mol= 2.9 moles of KMnO4

7) mass of erythromycin= 1.00×10^0 g

Molar mass of erythromycin= 733.93 g/mol

Number of moles of erthromycin= 1.00×10^0 g/ 733.93 g/mol = 1.36×10^-3 moles

8) mass of cortisone= 10.0 mg= 10×10^-3g

Molar mass of cortisone= 360.45 g/mol

Number of moles of cortisone= mass/molar mass= 10×10^-3 g/ 360.45 g/mol = 2.8×10^-5 moles

3 0

3 years ago

3.10 mol of an ideal gas with CV,m=3R/2 is expanded adiabatically against a constant external pressure of 1.00 bar. The initial

Otrada [13]

Answer : The value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

Explanation :

First we have to calculate the final temperature of the gas.

Using poission equation:

$(\frac{P_1}{P_2})^{1-\gamma}=(\frac{T_2}{T_1})^{\gamma}$

where,

$P_1$ = initial pressure = 14.7 bar

$P_2$ = final pressure = 1.00 bar

$T_1$ = initial temperature = 305 K

$T_2$ = final temperature = ?

$\gamma=\frac{R}{C_v}+1=frac{R}{(\frac{3R}{2})}+1=\frac{5}{3}=1.67$

Now put all the given values in the above expression, we get:

$(\frac{14.7}{1.00})^{1-1.67}=(\frac{T_2}{305})^{1.67}$

$T_2=103.7K$

Now we have to calculate the q, w, ΔU and ΔH for the process.

As we know that, at adiabatic process the value of q=0. So,

$dq=dU-w\\\\0=dU-w\\\\dU=w\\\\w=dU=n\times C_v\times (T_2-T_1)$

$w=dU=3.10mol\times \frac{3R}{2}\times (103.7-305)K$

$w=dU=3.10mol\times \frac{3\times 8.314J/mol.K}{2}\times (103.7-305)K$

$w=dU=-7782.278J=-7.78kJ$

Now we have to calculate the value of $C_p$

$C_p-C_v=R\\\\C_p=R+C_v\\\\C_p=R+\frac{3R}{2}\\\\C_p=\frac{5R}{2}$

Now we have to calculate the value of ΔH.

$dH=n\times C_p\times (T_2-T_1)$

$dH=3.10mol\times \frac{5R}{2}\times (103.7-305)K$

$dH=3.10mol\times \frac{5\times 8.314J/mol.K}{2}\times (103.7-305)K$

$dH=-12970.5J=-12.97kJ$

Thus, the value of q, w, ΔU and ΔH for the process is, 0 J, -7.78 kJ, -7.78 kJ and -12.97 kJ respectively.

4 0

3 years ago

Farmers who raise cotton once used arsenic acid, H₃AsO₄, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with

Stels [109]

Explanation:

The reaction equation will be as follows.

$H_{3}AsO_{4} \rightleftharpoons H_{2}AsO^{-}_{4} + H^{+}$

Hence, the expression for $K_{a}$ is as follows.

$K_{a} = \frac{[H_{2}SO^{-}_{4}][H^{+}]}{[H_{3}AsO_{4}]}$

Let us assume that the concentration of both $[H_{2}AsO^{-}_{4}]$ and $[H^{+}]$ is x.

$2.5 \times 10^{-4} = \frac{x \times x}{0.5}$

x = 0.01118034

This means that the concentration of $[H^{+}]$ is 0.01118034.

Since, we know that the relation between pH and concentration of hydrogen ions is as follows.

pH = $-log [H^{+}]$

= $-log (0.01118034)$

= 1.958

Thus, we can conclude that the pH of a 0.500 M solution of arsenic acid is 1.958.

6 0

4 years ago