How do you make clorifil?

What do all decomposition reactions have in common? (My answer: D, correct?)

inn [45]

Correct answer is D.

,,They start with one reactant and end with more than one product."

:-) ;-)

3 0

3 years ago

Read 2 more answers

The synthesis of nitrogen trihydride from nitrogen gas and hydrogen gas is shown by which balanced chemical equation?

Bumek [7]

The correct answer is: [B]:
___________________________________________________________
" N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) " .
___________________________________________________________
Note of interest:
___________________________________________________________
This particular reaction is known as the "Haber process" .
___________________________________________________________

7 0

3 years ago

Read 2 more answers

How does the density of a gas depend on the molar mass of the gas?

AlladinOne [14]

Answer:

The density of the ideal gas is directly proportional to its molar mass.

Explanation:

Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

$\rho = \frac{m}{V}$ ......equation (1)

According to the ideal gas law for ideal gas:

$PV = nRT$ ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

As we know,

The number of moles: $n = \frac{m}{M}$

where m is the given mass of gas and M is the molar mass of the gas

So equation (2) can be written as:

$PV = \frac{m}{M}RT$

⇒ $PM= \frac{m}{V} RT$

⇒ $\frac{PM}{RT}= \frac{m}{V}$ ......equation (3)

Now from equation (1) and (3), we get

$\frac{PM}{RT}= \frac{m}{V} = \rho$

⇒ Density of an ideal gas: $\rho = \frac{PM}{RT}$

⇒ Density of an ideal gas: ρ ∝ molar mass of gas: M

Therefore, the density of the ideal gas is directly proportional to its molar mass. 

6 0

3 years ago

What is the SI base unit for length?

sp2606 [1]

The SI base unit for length is meter.

In order to make smaller measurements, you can use the centi-, milli-, micro-, etc. prefixes.

When you want to reference larger measurements, you can use the kilo-, mega-, giga- and prefixes such as those.

5 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