What salt is produced from mixing csoh and h2co3

The mass of magnesium consumed amounts to 21.42g. Explanation: As per the balanced reaction, three moles of magnesium react with one mole of nitrogen, producing one mole of magnesium nitride. Given that the mass of nitrogen involved equals 8.33g, the moles of the reacted nitrogen equal a particular quantity. The required moles of magnesium, therefore, equals three times the moles of nitrogen utilized. Subsequently, the mass of magnesium needed is determined by multiplying the moles by its molar mass.

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

The half-reactions are:

Anode oxidation half reaction:  

Cathode reduction half reaction:  

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature =

n = electrons exchanged in oxidation-reduction = 2

= standard electrode potential for the cell = +0.63 V

= cell potential for the reaction =?

= concentration of Zn²⁺ = 3.5 M

=

Now substituting all known values into the equation, we arrive at:

So, the resulting cell potential for this reaction is 0.50 V

The formula for Molarity is given by:

                                  M = moles / V
To isolate V,
                              V = moles / M ------------------(1)
Moles can also be calculated as:
                                  moles = mass / M.mass -------------(2)
Substituting the value of moles from equation 2 into equation 1 yields:
                                  V = (mass / M.mass) / M
Plugging in the numbers gives:
                                  V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³.

Answer:

Ir(NO2)3

Explanation:

The molar mass is 330.2335, in case that's also required.

Convert HCl and H2O to moles.

36.0 g of HCl = 0.987 moles HCl

98.0 g of H2O = 5.44 moles H2O

Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.

For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.

Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O

Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.

11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O

11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.