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2 years ago

What are five different types of air pollutant each

1 answer:

7 0

5 types of air pollution would include: Carbon Monoxide, Lead, Nitrogen oxides, sulfur oxides. I only know those 4 actually… hope it helped

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About 3% of the water on Earth is freshwater. Only about 40% of that freshwater is available for human use. Why is so much fresh

Ilya [14]

I believe the answer you're looking for is B. It's Polluted. I hope this helps.

3 0

2 years ago

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Please help. Thank you!

scoray [572]

Answer: Uranium is a chemical element with atomic number 92 which means there are 92 protons and 92 electrons in the atomic structure. The chemical symbol for Uranium is U. Electron configuration of Uranium is [Rn] 5f3 6d1 7s2. Possible oxidation states are +3,4,5,6.

Explanation:

6 0

2 years ago

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Hydrogenation reactions, in which H2 and an "unsaturated" organic compound combine, are used in the food, fuel, and polymer indu

ddd [48]

Answer: The amount of heat released is 56 MJ.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $C_2H_6$ = 12 kg = 12000 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $C_2H_6$ = 30 g/mol

Putting values in above equation, we get:

$\text{Moles of }C_2H_6=\frac{12000g}{30g/mol}=400mol$

The chemical reaction for hydrogenation of ethene follows the equation:

$C_2H_4+H_2\rightarrow C_2H_6$

By Stoichiometry of the reaction:

When 1 mole of ethane releases 140 kJ of heat.

So, 400 moles of ethane will release = $\frac{140}{1}\times 400=56000kJ$ of heat.

Converting this into Mega joules, using the conversion factor:

1 MJ = 1000 kJ

So, $\Rightarrow 56000kJ\times (\frac{1MJ}{1000kJ})=56MJ$

Hence, the amount of heat released is 56 MJ.

8 0

2 years ago

Gifblaar is a small South African shrub and one of the most poisonous plants known because it contains fluoroacetic acid (FCH2CO

PSYCHO15rus [73]

[H_{3}O^{+}] = 0.00770 M

The equilibrium equation representing the dissociation of $FCH_{2}COOH$

$FCH_{2}COOH(aq) + H_{2}O (l) FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq)$

Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

So y = $[H_{3}O^{+}]$ = $[FCH_{2}COO^{-}]$ = 0.00770 M

$pK_{a} = 2.59K_{a} = 10^{-2.59} = 0.00257 M$

$K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}$

$0.00257 = \frac{0.00005929}{x - 0.00770}$

0.00257 x - 0.00001979 = 0.00005929

x = 0.031 M

Therefore, initial concentration of the weak acid is 0.031 M

4 0

3 years ago

Which one of the following is not a homogeneous mixture?

Kitty [74]

Vegetable soup should be the correct answer. The reason for this is that it can easily be split, where as the rest would be very difficault if not imposible because of chemical change.

3 0

3 years ago