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timurjin [86]
3 years ago
9

Complete combustion of 6.4g of compound K produced 8.8 g of carbon dioxide and 7.2 g of water. Calculate the empirical formula o

f K. ​
Chemistry
1 answer:
Elenna [48]3 years ago
7 0
Lh (x1)78.009+1= 0.0000567
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What happens to the volume of a gas in a closed container if the temperature increases, but the pressure remains the same? Why?
labwork [276]

Answer:

Volume will goes to increase.

Explanation:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.

Now we will put the suppose values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 4.5 L × 348 K / 298 k

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

Hence prove that volume increase by increasing the temperature.

3 0
3 years ago
What process describes the transfer of heat through matter by molecular activity?
DIA [1.3K]
The process that describes the transfer of heat through matter by activity of the molecules would be conduction. This type of heat transfer due to the motion of electrons and ions. within a body. As molecules collide with each energy is transferred and released which cause temperature to rise or drop.<span />
8 0
4 years ago
Read 2 more answers
I´LL GIVE BRAINLIEST ORDER THE FOLLOWING LIQUIDS FROM MOST DENSE TO LEAST DENSE: WATER, PANCAKE SYRUP, COOKING OIL, MILK, DISH S
NeX [460]

Answer:

sope syrup cooking oil milk water

Explansation:

5 0
3 years ago
These models show the electron structures of two different nonmetal elements.
Soloha48 [4]

Answer: Option (D) is the correct answer.

Explanation:

Valence shell is the shell present on the outermost core of an atom and electrons present in the valence shell are known as valence electrons.

If an atom has completely filled valence shell then it means the atom is not reactive in nature because it is already stable.

But when an atom has less than eight electrons in its valence shell then it means to attain stability the atom will readily attract electrons towards itself.

As the given element 1 has 8 electrons in its valence shell. Hence, it is not reactive in nature but element 2 has 6 valence electrons. So, in order to attain stability element 2 will readily attract 2 electrons from a donor atom.

Thus, we can conclude that element 2 is more reactive because it does not have a full valence shell, so it will attract electrons.

4 0
3 years ago
Whats the voltage of CuCl2 + Zn -&gt; ZnCl2 + Cu
gtnhenbr [62]

Answer:

Approximately 1.10\; {\rm V} under standard conditions.

Explanation:

Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

Write down the ionic equation for this reaction:

\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
  • Copper gains electrons and was reduced (at the cathode): {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

7 0
2 years ago
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