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Lubov Fominskaja [6]

1 year ago

7

The _ atom in a water molecule points toward the sodium ion because its partial _ charge is attracted to the sodium

ion's ______ charge. This is called _______.

1 answer:

makvit [3.9K]1 year ago

8 0

Answer:

Below

Explanation:

Oxygen negative positive hydration

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What is woven into farbic?

LenaWriter [7]

Woven fabric is any textile formed by weaving. Woven fabrics are often created on a loom, and made of many threads woven on a warp and a weft. Technically, a woven fabric is any fabric made by interlacing two or more threads at right angles to one another.

4 0

3 years ago

A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4

natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

$M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M$

8 0

3 years ago

Lagta ha bhaijaan sa sab bahenjaan naraz haan koi bat ni kr raha

Fiesta28 [93]

Answer:

maybe,lol

Explanation:

5 0

2 years ago

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

Thus, the anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

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

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

The overall cell reaction will be,

$2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}$

$E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V$

$E^o_{[Cu^{2+}/Cu]}=0.34V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

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

$E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=1.54V-(0.34V)=1.20V$

Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}$

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:

$E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}$

$E_{cell}=1.022V$

Therefore, the emf of cell potential is 1.022 V

5 0

3 years ago

Which contains more molecules: 340g of NH3 or 90g of H2O show your working!!

Shtirlitz [24]

Answer:

atomic mass ( N =14 , H = 1×3 that means 14+3=17

340gram×17gram / mole

=5780mole

atomic mass ( H= 1×2 ,O= 16 =16+2 =18

90gram×18gram /mole

=1620mole

I think 340g ammonia (NH3)

6 0

3 years ago

Read 2 more answers