Please help me it's just one question!

How many formula units make up 10.2 g of magnesium chloride (MgCl2)?

beks73 [17]

Answer:

6.46×10²² formula units

Explanation:

From the question given above, the following data were obtained:

Mass of MgCl₂ = 10.2 g

Number of formula units =?

From Avogadro's hypothesis,

1 mole of MgCl₂ = 6.02×10²³ formula units

But,

1 mole of MgCl₂ = 24 + (35.5×2) = 24 + 71 = 95 g

Thus, we can say:

95 g of MgCl₂ = 6.02×10²³ formula units

Finally, we shall determine the formula units in 10.2 g MgCl₂. This can be obtained as follow:

95 g of MgCl₂ = 6.02×10²³ formula units

Therefore,

10.2 g of MgCl₂ = (10.2 × 6.02×10²³) / 95

10.2 g of MgCl₂ = 6.46×10²² formula units

Thus, 10.2 g of MgCl₂ contains 6.46×10²² formula units

6 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

Hi! I need some solar system facts!

nikitadnepr [17]

The Solar System[b] is the gravitationally bound system of the Sun and the objects that orbit it, either directly or indirectly.[c] Of the objects that orbit the Sun directly, the largest are the eight planets,[d] with the remainder being smaller objects, the dwarf planets and small Solar System bodies. Of the objects that orbit the Sun indirectly—the natural satellites—two are larger than the smallest planet, Mercury.[e]

5 0

3 years ago

Read 2 more answers

E is solution of HCI of unk concentration f contain 4.8g of NaOH in 25cm point of salution

Readme [11.4K]

Answer:

18

Explanation:

no explanation sorry

5 0

3 years ago

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/

lara31 [8.8K]

Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

2.67%

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

You are calculating weight /Volume not weight/weight % or percent by mass of the solute

Here the weight/weight % or percent by mass of the solute is asked : So first convert the VOLUME OF SOLUTION into MASS

Density of potassium nitrate in water KNO3 = 2.11 g/mL

$density=\frac{mass}{volume}$