How is molarity measured?

La formula quimica de la molecula de agua H20 , si H = 1 gramos y O = 16 gramos . Cual es su composicion porcentual ? 88,88 % de

vesna_86 [32]

Answer:

88,88 % de O y 11,11 % de H

Explanation:

La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:

<em>Masa molar agua:</em>

2H = 2*1g/mol = 2g/mol

1O = 1*16g/mol = 16g/mol

Masa molar = 2 + 16 = 18g/mol

100g H2O * (1mol / 18g) = 5.556 moles H2O.

Moles de hidrógeno:

5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H

Moles Oxígeno = Moles H2O = 5.556 moles

La masa de hidrógeno es:

11.11mol * (1g/mol) 11.11g H

La masa de oxígeno es:

5.556 mol * (16g / 1mol) = 88.89g O

Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:

7 0

2 years ago

How can there be more than 1000 different atoms when there are only about 100 different elements?

Irina-Kira [14]

Um im pretty sure there are only about 100 different atoms...

7 0

2 years ago

If 250.0 mL of a 0.96 M solution of acetic acid (C 2H 4O 2) are diluted to 800.0 mL, what will be the approximate molarity of t

adoni [48]

Answer:

0.30M HOAc

Explanation:

Given 250.0ml (0.96M HOAc) => 800ml(??M HOAc)

Use the dilution equation...

(Molarity x Volume)concentrated soln = (Molarity x Volume)dilute soln

(0.96M)(250.0ml) = (Molarity diluted soln)(800ml)

Molarity diluted soln = (0.96M)(250.0ml)/(800ml) = 0.30M HOAc

7 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

Epsom salts are hydrates of magnesium sulfate. the formula for epsom salt is mgso4·7h2o. if a 7.834 g sample is heated until con

vesna_86 [32]

1. For the first question, we must find the mass of the anhydrous salt, MgSO₄. The molar mass for MgSO₄·7H₂O is 246.47 g/mol, while that of MgSO₄ is 120.37 g/mol.

7.834 g MgSO₄·7H₂O * 1 mol MgSO₄·7H₂O/246.47 g * 1 mol MgSO₄/1 mol MgSO₄·7H₂O * 20.37 g MgSO₄/mol = <em>0.647 g MgSO₄</em>

2. Mass of Water = Mass of sample - Mass of anhydrous sale
Mass of water = 7.834 - 0.647 = 7.19 g
Percent Water in Hydrate = 7.19/7.834 * 100 = <em>91.74%</em>

5 0

3 years ago