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3 years ago

Is all air pollution caused by humans

Chemistry

2 answers:

NeX [460]3 years ago

5 0

Not all air pollution is caused by humans but I’d say at least a grand total of 90% of it is at my best guess!

larisa86 [58]3 years ago

4 0

Well partly yes because air pollution can happen when we burn fossil fuels, it can happen when we use vehicles and household/farming chemicals can cause air pollution so yes it is caused by humans

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What volume of a 2.25 M sodium chloride solution will contain 4.58 moles of sodium chloride

Rzqust [24]

Answer:

Option E. 2.04 L

Explanation:

Data obtained from the question include:

Molarity of NaCl = 2.25 M

Mole of NaCl = 4.58 moles

Volume =..?

Molarity is simply defined as the mole of solute per unit litre of the solution. It is represented mathematically as:

Molarity = mole /Volume

With the above formula, we can obtain the volume of the solution as follow:

Molarity = mole /Volume

2.25 = 4.58/volume

Cross multiply

2.25 x volume = 4.58

Divide both side by 2.25

Volume = 4.58/2.25

Volume = 2.04 L

Therefore, the volume of the solution is 2.04 L

6 0

2 years ago

Answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)

Natali [406]

Answer:

a) 1.44g

b) 77.3%

Explanation:

1. Chemical balanced equation (given)

$NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)$

2. Mole ratio

$1molNaOH(aq):1molKHP(s)$

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

3. Find the number of moles in 72.14 mL of the base

$Molarity=\text{number of moles of solute}/\text{volume of solution in liters}$

$\text{Volume of solution}=72.14mL=0.07214liters$

$\text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol$

4. Find the number of grams of KHP that reacted

The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

Convert moles to grams:

mass = number moles × molar mass = 0.007055mol × 204.23g/mol

mass = 1.4408 g.

You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

5. Find the percentage of KHP in the sample

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

% = (mass of substance / mass of sample) × 100

% = (1.4408g/ 1.864g) × 100 = 77.3%

7 0

2 years ago

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

3 years ago

What is the product for MgBr2+ Cl2

nikklg [1K]

???...........…..............

5 0

3 years ago

Compare the trends for atomic size and first ionization energy.

bazaltina [42]

Explanation:

On a periodic table, the atomic size is depicted by the radius of the atom.

Across a period, the atomic radius reduces progressively from left to right.

Down the group from top to bottom, atomic radii increases progressively.

For the ionization energy, from left to right, across the period, it increases progressively and down a group it reduces.

These two trends are related in that as the atomic radius decreases across the period there is an increasing nuclear charge which is not compensated for by the the successive shells of electrons being added. This also similar down the group.

6 0

2 years ago