All categories

3 years ago

What is the third most common gas found in the air we breath?

2 answers:

6 0

Explanation: The most common gas people breathe from clean air is nitrogen. Air is a mixture of gases, and the air on Earth contains about 78 percent nitrogen and about 21 percent oxygen. Argon, carbon dioxide neon, helium, methane, krypton, hydrogen and xenon are in much smaller amounts.

Setler [38]3 years ago

6 0

Answer: Carbon dioxide

Explanation:

You might be interested in

The specific heat of zinc is 0.39 J/g*°C. How much energy needed to change the temperature of 34g of zinc from 22°C to 57°C. Is

Svetlanka [38]

Answer:

464.1 J absorbed.

Explanation:

Given data:

Specific heat of zinc = 0.39 J/g°C

Mass of zinc = 34 g

Temperature changes = 22°C to 57°C

Energy absorbed or released = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57°C - 22°C

ΔT = 35°C

Q = m.c. ΔT

Q = 34 g. 0.39 J/g°C. 35°C

Q = 464.1 J

6 0

3 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

You have 150. g of a bleach solution. The percent by mass of the solute, sodium hypochlorite (NaOCI), is 3.62%.

Taya2010 [7]

Answer:

There are 5.43 grams of NaOCl

Explanation:

The given percent by mass of the solute tells us that out of the 150 g of the solution, 3.62% are due solely to the solute.

In other words, the mass of the solute in the solution is:

150 g * 3.62/100 = 5.43 g

Thus, in 150 grams of the given bleach solution, there are 5.43 grams of sodium hypochlorite.

4 0

2 years ago

Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

saveliy_v [14]

Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; 
At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; 
so 0.17 = x² / (0.20 - x); 
Solving for x using the quadratic formula: 
x = [H+] = 0.063 M or pH = - log [H+] = 1.2.

7 0

3 years ago

Why is it necessary to use models to study submicroscopic objects such as atoms and molecules

Wittaler [7]

Answer:

It is necessary to use models to study sub- microscopic objects such as atoms and molecules because they are too small to be seen.

6 0

3 years ago