What is thermal energy?

Which of these are formed when

makvit [3.9K]

Option d lo siento si es incorrecto

7 0

3 years ago

Plz help :(

ziro4ka [17]

<h2>Answer:0.5 moles,1 mole</h2>

Explanation:

$Sr(NO_{3})_{2}$ is readily soluble in water.

$Sr(NO_{3})_{2}$ dissolves in water to give $Sr^{2+}$ ions and $NO_{3}^{-}$ ions.

The reaction is

$Sr(NO_{3})_{2}$ → $Sr^{2+}$ $+2NO_{3}^{-}$

So, $1$ mole of $Sr(NO_{3})_{2}$ gives $1$ mole of $Sr^{2+}$ and $2$ moles of $NO_{3}^{-}$ .

So, $0.5$ moles of $Sr(NO_{3})_{2}$ gives $0.5$ moles of $Sr^{2+}$ and $1$ mole of $NO_{3}^{-}$ .

5 0

4 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at 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:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

The combination of potassium-sparing diuretics and salt substitutes can result in dangerously high blood levels of:

alisha [4.7K]

Answer:

b. potassium.

Explanation:

Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.

Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.

4 0

3 years ago

g Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no r

Mrac [35]

Answer:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

Explanation:

We identify the reactants:

CuBr₂ and Pb(CH₃COO)₂

The products will be: Cu(CH₃COO)₂ and PbBr₂

You may know these information:

Salts from acetate are soluble.

Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺

PbBr₂ is formed, so this will be our precipitate

The equation is:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

8 0

3 years ago