A 2 mole sample of F2(g) reacts with excess NaOH(aq) according to the equation above. If the reaction is repeated with excess NaOH(aq) but with 1 mole of F2(g), which of the following is correct? Group of answer choices

The amount of OF2(g) produced is doubled.

The amount of OF2(g) produced is halved.

The amount of NaF(aq) produced remains the same.

The amount of NaF(aq) produced is doubled.

When instead of 2 moles of F₂, 1  mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.

Let's consider the balanced reaction that occurs when F₂ reacts with NaOH.

2 F₂(g) + 2 NaOH(aq) → OF₂(g) + 2 NaF(aq) + H₂O(l)

First, let's see the moles of OF₂ and NaF obtained from 2 moles of F₂, using the molar ratios derived from the balanced chemical equation.

Now, let's compare with the moles of OF₂ and NaF obtained from 1 mol of F₂, again using the same molar ratios derived from the balanced chemical equation.

As we can see, since we have half the amount of F₂, we obtain half the amount of the products. Then, the only right option is: The amount of OF₂(g) produced is halved.

When instead of 2 moles of F₂, 1  mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.

You can learn more about molar ratios here:

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

1 a

2 d

3 a

4 c

explanation: