Which describes two phases in dynamic equilibrium?

What is the empirical formula of a compound that is 18.8% Li, 16.3% C , and 64.9% O

zlopas [31]

Greetings!

To find the empirical formula you need the relative atomic mass of each element!

Li = 6.9

C = 12

O = 16

You can simply change the percentages into full grams

Li = 18.8g

C = 16.3g

O = 64.9

Then you use this to find the Number of moles = amount in grams / atomic mass

Li = 18.8 ÷ 6.9 = 2.7246

C = 16.3 ÷ 12 = 1.3583

O = 64.9 ÷ 16 = 4.0562

Then divide each number of moles by the smallest value:

Li = 2.7246 ÷ 1.3583 = 2.0

C = 1.3583 ÷ 1.3583 = 1

O = 4.0562 ÷ 1.3583 = 2.9 ≈ 3

So that means that there are 2 Li, 1 C, and 3 O

Empirical formula would be:

Li₂CO₃

Hope this helps!

3 0

4 years ago

Determine the volume in mL of 0.242 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 46.79

Alla [95]

The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL

<h3>Balanced equation </h3>

CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O

From the balanced equation above,

The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
The mole ratio of the base, NaOH (nB) = 1

<h3>How to determine the volume of NaOH</h3>

Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
Molarity of base, NaOH (Mb) = 0.242 M

Volume of base, KOH (Vb) =?

MaVa / MbVb = nA / nB

(0.204 × 46.79) / (0.242 × Vb) = 1

Cross multiply

0.242 × Vb = 0.204 × 46.79

Divide both side by 0.242

Vb = (0.204 × 46.79) / 0.242

Vb = 39.44 mL

Thus, the volume of NaOH needed for the reaction is 39.44 mL

Learn more about titration:

brainly.com/question/14356286

5 0

2 years ago

What makes hydrogen peroxide a polar molecule?

kupik [55]

<span>C. polar bonds and asymmetrical structure
If the molecule contains polar bonds but it has a symmetrical structure, the polar bonds will cancel each other out so the overall molecule will be non-polar.
On the other hand, if the molecule contains polar bonds but has an asymmetrical structure, then the polar bonds won't cancel each other out, so the overall molecule ends up being polar.
</span>

7 0

3 years ago

Read 2 more answers

Write the oxidation and reduction half reactions;

luda_lava [24]

Answer:

a)

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

b)

$Mg$ ⇒ $Mg^{2+}+2e^-$

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

3 0

4 years ago

Read 2 more answers

7.66 Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water,

White raven [17]

Explanation:

(a) potassium oxide with water

$K_2O(s)+H_2O(l)\rightarrow 2KOH(aq)$

According to reaction,1 mole of potassium oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.

(b) diphosphorus trioxide with water

$P_2O_3(s)+3H_2O(l)\rightarrow 2H_3PO_3(aq)$

According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water to give 2 moles of phosphorus acid.

(c) chromium(III) oxide with dilute hydrochloric acid,

$Cr_2O_3(s)+6HCl(aq)\rightarrow 2CrCl_3(aq)+3H_2O(l)$

According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.

(d) selenium dioxide with aqueous potassium hydroxide

$SeO_2 (s)+2KOH (aq)\rightarrow K_2SeO_3(aq)+H_2O(l)$

According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.

6 0

4 years ago