Why should carbonated beverages be kept cold?

These substances are all ____. H2O2 Li NaCl O2

Bogdan [553]

Answer: pure substances.

Explanation:

The given substances are:

H₂O₂
Li
NaCl
O₂

All what surrounds us, which has mass and occupies spaces, is matter. There are two kind of matter: pure substances and mixtures.

Pure substances have a uniform and constant composition. On the other hand, mixtures are combinations of two or more pure substances in any arbitratry ratio.

Pure substances may be elements or compounds. The elements are the substances conmposed by one only kind of atom. In the list of substances given, Li and O₂ are elements: all the atoms in Li are lithium, and all the atoms in O₂ are oxygen atoms.

Compounds are the chemical combination of two or more different kind of atoms. In the given list H₂O₂ and NaCl are compounds. As you see, H₂O₂ contains atoms of hydrogen and oxygen, chemically bonded, in a fixed ratio (2 atoms of hydrogen by 2 atoms of oxygen). And NaCl has atoms of Na (sodium) and Cl (chlorine), chemicaly bonded, in a fixed ratio (1:1).

There are only 118 known elements and you can find them in any modern periodic table. Therer are virtually infinitely many compounds since many different combinations of the elements can be attained.

Elements and compounds have in common that they are classified as pure substances.

7 0

3 years ago

The boiling point of a substance is tested. After 10 tests, the result is given as 37+/−3°C. Which conclusion can be drawn from

VMariaS [17]

Answer:

Explanation:

The actual boiling point is probably between 34C and 40C.

6 0

2 years ago

Read 2 more answers

What would be the most likely explanation if none of the tested paint samples had turned brown?

mariarad [96]

Answer:4/d

Explanation:

i did this before

7 0

2 years ago

If a catalyst is used in a reaction,

Ann [662]

If a catalyst is used in a reaction, C. The reaction rate increases.

4 0

3 years ago

Whats the voltage of CuCl2 + Zn -> 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

1 year ago