Identify the limiting reactant when 1.22 g of O2 reacts with 1.05 g H2 to produce water.

Answer:

yes

Explanation:

yes

Answer:

To achieve the desired outcome, 8.55 mL of NaOH is necessary.

Explanation:

Considering that:

the weak acid has a mass of 0.4 g

and a molecular weight of 234 g/mol

so, the number of moles of the weak acid is calculated as 0.4 g/234 g/mol = 0.00171 mole

To convert half of the weak acid (WA) to conjugate base (CB), we must add NaOH.

Thus, [WA]=[CB] 0.00171/2 = 8.55×10⁻⁴ mole of NaOH required

Further, knowing that the concentration of NaOH is 0.10 M

the volume needed to achieve this result can be calculated as follows:

=

= 8.55 mL of NaOH is necessary

Answer:

To lower the temperature of the solution from 25.0°C to 5.0°C, it is necessary to use 35.2g of NH₄NO₃ for every 100.0g of water.

Explanation:

In order to cool down the solution, we need:

4.184 J/g°C × (5.0°C - 25.0°C) × (100.0g + X) = -Y

8368 J + 83.68 J/gX = Y (1)

Here, x represents the grams of NH₄NO₃ required, and Y represents the energy needed to remove heat.

Furthermore, the energy Y becomes:

Y = 25700 J/mol × X

Y = 321 J/g X (2)

Substituting (2) into (1)

8368 J + 83.68 J/g X = 321 J/g X

8363 J = 237.32 J/gX

X = 35.2g

This means 35.2g of NH₄NO₃ must be used for every 100.0g of water to achieve a temperature decrease from 25.0°C to 5.0°C.

I trust this information will be useful!