Which reaction creates more massive nuclei?

What is the molar mass of Ca(NO3)2?

gregori [183]

Answer:

C.164.1g/mol

Explanation:

40+14×2+16×6

164

7 0

2 years ago

PLEASE HELP ME !!!!!!!

love history [14]

Answer:

it would be 5,045

Explanation:

because it is closer to 5,000. pls correct me if wrong

5 0

2 years ago

How could you “supersaturate” solutions, exceeding the amount of dissolved solute possible

Bingel [31]

Answer:

Explanation:

In a saturated solution, more solute cannot be dissolved at a given temperature.

This is because, the solute dissolves in a solvent because of space between particles of solvent but on continuous addition of solute, the space between the solvent particles gets fulfilled. Thus no more solute particle can dissolve in a solvent.

4 0

1 year 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

2 years ago

How are compounds Similar to elements and how are they different

Yuki888 [10]

Answer:

A compound contains atoms of different elements chemically combined together in a fixed ratio. An element is a pure chemical substance made of same type of atom. Compounds contain different elements in a fixed ratio arranged in a defined manner through chemical bonds. They contain only one type of molecule.

3 0

2 years ago