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3 years ago

14

[PLEASE ANSWER FAST] : When Sodium reacts with oxygen, it gives Sodium oxide. It is a redox reaction. What is the substance redu

ced here?

2 answers:

kondaur [170]3 years ago

6 0

Answer:

The substance reduced is oxygen.

Hope this helps.

Tresset [83]3 years ago

4 0

Answer:

Oxygen is reduced

Explanation:

This is because when oxygen is changing to Sodium oxide , metal is added . The addition of the metal there is reduction in this sense oxygen is reduced but with the sodium , oxygen is being added to the sodium hence sodium is oxidized

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Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles$

b) moles of $I_2$

$\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles$

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

According to stoichiometry :

1 mole of $I_2$ require 1 mole of $H_2$

Thus 0.0787 moles of $l_2$ require= $\frac{1}{1}\times 0.0787=0.0787moles$ of $H_2$

Thus $l_2$ is the limiting reagent as it limits the formation of product and $H_2$ acts as the excess reagent. (10.0-0.0787)= 9.92 moles of $H_2$ are left unreacted.

Mass of $H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g$

Thus 19.9 g of $H_2$ remains unreacted.

5 0

3 years ago

If a student mixes 75 mL of 1.30 M HNO3 and 150 mL of 6.5 M NaOH. is the final solution acidic, basic, or neutral

raketka [301]

Answer:

The solution is basic.

Explanation:

We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):

If no. of millimoles of acid > that of base; the solution is acidic.

If no. of millimoles of acid = that of base; the solution is neutral.

If no. of millimoles of acid < that of base; the solution is basic.

We need to calculate the no. of millimoles of acid and base:

no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.

no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.

<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>

<em>So, the solution is: basic.</em>

4 0

3 years ago

What is the approximate tilt of the earth?<br> 18°<br> 23°<br> 45°<br> 60°

balandron [24]

The answer is 23.5° but I guess 23° is closest

7 0

3 years ago

Read 2 more answers

A car is traveling at 15kph and speeds up to 30kph in 5 seconds. What is its acceleration?

Wewaii [24]

Answer:

the acceleration is 3 k/h/s

Explanation:

The formula for acceleration is Δv/t, so change in velocity divided by time.

The original velocity is 15 kph, and the final velocity is 30 kph. Therefore, the change in velocity is 15 kph. 15/5=3 k/h/s

3 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

4 years ago