The reaction between $m_{2}SO_{4}$ with calcium chloride can be shown as- $m_{2}SO_{4}$ + $CaCl_{2}$ → $CaSO_{4}$ ↓+2mCl. The molecular weight of $CaSO_{4}$ is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of $m_{2}SO_{4}$ = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 moles $m_{2}SO_{4}$ to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989× $10^{-3}$ moles of calcium sulfate.

Thus, 9.989× $10^{-3}$ moles of $m_{2}SO_{4}$ reacts in this reaction.

Let assume the atomic mass of m is x thus the molecular weight of $m_{2}SO_{4}$ is 2x+96.

So we may write 9.989× $10^{-3}$ × (2x+96) =1.42

Or, 2x + 96 = 142.146

Or, 2x = 46.146

Or, x = 23.073

Thus the atomic mass of m is 23.073. The atom (m) is sodium (Na).

8 0

2 years ago

According to the theory of plate tectonics,

11Alexandr11 [23.1K]

A I’m sure of it because it only makes since one would think ✊

3 0

2 years ago

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

Lilit [14]

Answer:

pH = 1.32

Explanation:

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

3 0

2 years ago

Write a balanced equation from each cell notation:(b) Fe(s) | Fe²⁺(aq) | NO₃⁻(aq) | NO(g) | Pt(s)

tatiyna

The balanced redox reaction is

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

<h3>What is Galvanic cell? </h3>

It is a device that is used for the conversion of the chemical energy which is produce in the redox reaction into the electrical energy. It is also termed as the voltaic cell or electrochemical cell.

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

<h3>What is redox reaction? </h3>

The reaction in which both reduction and oxidation reaction takes place simultaneously is known as redox reaction.

<h3>What is oxidation reaction? </h3>

The reaction in which a substance or compound or species looses its electrons. In this reaction, oxidation state of an element increases.

<h3>What is reduction reaction? </h3>

The reaction in which a substance or compound or species accept the electrons. In this reaction, oxidation state of an element decreases.

The given two-half reactions are:

At anode:

3Fe → 3Fe²⁺ + 6e-

At cathode:

8H(+) + 2NO₃⁻ → 2NO + 4H2O

Redox reaction:

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

According to the voltaic cell, the half-cell reaction (1) shows oxidation reaction therefore, it occurs at anode and the half-cell reaction (2) shows reduction reaction therefore, it occurs at cathode.

Thus, we concluded that the redox reaction is

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

learn more about voltaic cell:

brainly.com/question/4052514

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7 0

1 year ago