What is the function of the excretory system? PLEASE HELPP

What are atoms?

snow_lady [41]

Answer:

a. tiny particles that make up all matter

6 0

3 years ago

Read 2 more answers

Define the term ionization energy. Choose one:

olga nikolaevna [1]

Answer:

A. The amount of energy needed to remove 1 mole of electrons from 1 mole of ground-state atoms or ions in the gas phase.

Explanation:

Ionization energy is the quantity of energy required to remove an electron in ground electronic state from an isolated gaseous atom or ion, resulting in a cation. kJ/mol is the expresion we use for this energy, it refers to the amount of energy it takes for all the atoms in a mole to lose one electron each.

Ionization energy can be used as an indicator of reactivity and can be used to help predict the strength of chemical bonds because the more electrons are lost, the more positive the ion will be and the harder it will be to separate the electrons from the atom.

I hope you find this information useful and interesting! Good luck!

8 0

3 years ago

Was the Soviet Union's use of the Aral Sea

Ivan

Answer:

Yes it was

Explanation:

It was broken because they did not know what they were doing

4 0

3 years ago

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

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

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

3 years ago

Im hvaing a hard time getting the right answer

Tanya [424]

Answer:

$V=23.9mL$

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

$n=2.49g*\frac{1mol}{125.55 g}=0.0198mol$

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

$M=\frac{n}{V}\\\\V=\frac{n}{M}$

By plugging in the moles and molarity, we obtain:

$V=\frac{0.0198mol}{0.830mol/L}=0.0239L$

Which in mL is:

$V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL$

Best regards!

6 0

3 years ago