This is the chemical formula for nickel tetracarbonyl (a powerfully poisonous liquid used in nickel refining) Ni(CO)4 A chemical

Response:

702 mL

To elaborate:

Given the following:

Mass of sodium hydroxide = 13.20 g

Molarity of H₂SO₄ = 0.235 M

We're tasked with finding the volume of acid necessary to neutralize the sodium hydroxide solution

Step 1: Write the balanced reaction equation

The reaction between H₂SO₄ and NaOH can be summarized as follows:

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Step 2: Calculate the moles of NaOH

Moles are calculated by dividing mass by molar mass

The molar mass of NaOH is 40.0 g/mol

Hence;

Number of moles of NaOH = 13.20 g ÷ 40 g/mol

= 0.33 moles of NaOH

Step 3: Determine moles of H₂SO₄ that react

According to the balanced equation, 2 moles of NaOH react with 1 mole of H₂SO₄

Therefore, the ratio giving moles of H₂SO₄ = Moles of NaOH ÷ 2

= 0.33 moles ÷ 2

= 0.165 moles

Step 4: Find the volume of H₂SO₄

Molarity indicates the concentration of the solution in moles per liter

Molarity = Moles ÷ Volume

By rearranging the formula, we find volume = Moles ÷ Molarity

= 0.165 moles ÷ 0.235 M

= 0.702 L

= 702 mL

Thus, the volume of the 0.235 M H₂SO₄ acid solution required equals 702 mL

Answer:

81°C.

Justification:

We can arrive at this conclusion using the formula:

Q = m.c.ΔT,

where Q denotes the heat lost by water (Q = - 1200 J).

m represents the mass of water (m = 20.0 g).

c indicates the specific heat of water (c = 4.186 J/g.°C).

ΔT signifies the difference between the starting temperature and the final temperature (ΔT = final T - initial T = final T - 95.0°C).

Given Q = m.c.ΔT

It follows that (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

Consequently, the correct answer is: 81°C.

The reaction can be described as follows: CO + 2H2 = CH3OH. Given the specified quantities of the reactants, we will identify the limiting reactant and compute the remaining excess amount. Calculating, 1.50 x 10^-6 g CO converts to 5.36 x 10^-8 mol CO, while 6.80 x 10^-6 g H2 equals 3.37 x 10^-6 mol H2. Thus, CO is fully consumed in the reaction, leaving 3.37 x 10^-6 - 5.36 x 10^-8 = 3.32 x 10^-6 moles of gas.

In a water molecule, the sharing of electrons occurs between the oxygen and hydrogen atoms within covalent bonds; however, this sharing is unequal. The oxygen atom holds a stronger pull on the electrons compared to the hydrogen atoms in the bond.