Chlorine has 7 valence electrons. Hydrogen has 1 valence

What is the mass percent of a solution of 7.6 grams sucrose in 83.4 grams of water

dlinn [17]

Answer:

The mass percent of a solution of 7.6 grams sucrose in 83.4 grams of water is 8.351 %.

Explanation:

Given,

Mass of Sucrose = 7.6 grams

Mass of Water = 83.4 grams

In this solution, Sucrose is solute and water is the solvent.

Mass percent of a solution can be calculated using the formula,

Mass percent = (Mass of Solute/Mass of Solution)(100)

As sucrose is solute, mass of solute = 7.6 grams

As the solution contains both Sucrose and Water,

Mass of solution = 7.6 grams + 83.4 grams = 91 grams

Substituting the values, Mass percent = (7.6/91)(100) = 8.351 %.

7 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

A 10 gram sample of material, a, was placed in water. four grams of it, b, did not dissolve, even after being separated and plac

kicyunya [14]

Sksksk isjsjs jsjsjs jsjsjs jsjsjs jsjsjs ksjss

3 0

3 years ago

The fact that air pressure increases the closer you get to Earth is due to ________?

Rudiy27

Density that is determined by a balance between gravity and pressure

4 0

2 years ago

Sulfur burns in the air to form 33.6L of sulfur dioxide. (S = 32, O =16, 64g of sulfur dioxide has 22.4L, 32g of oxygen has 22.4

ivolga24 [154]

The mass of oxygen reacted/required in this reaction is obtained as 48g.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.

Now, we have the reaction; S + O2 ------>SO2

If 1 mole of sulfur dioxide contains 22.4 L

x moles of sulfur dioxide contains 33.6L

x = 1.5 moles of sulfur dioxide.

Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.

Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g

Learn more anout stoichiometry: brainly.com/question/9743981

8 0

2 years ago