Answer :
To determine how much substance [tex]\( AB \)[/tex] Colin would make, we'll follow these detailed steps:
1. Identify the given quantities of the reactants:
- Colin added 10 grams of substance [tex]\( A \)[/tex].
- Colin also added 45 grams of substance [tex]\( B \)[/tex].
2. Understand the reaction:
- The reaction is given by [tex]\( A + B \rightarrow AB \)[/tex], indicating a 1:1 molar ratio. This implies that one mole of [tex]\( A \)[/tex] reacts with one mole of [tex]\( B \)[/tex] to form one mole of [tex]\( AB \)[/tex].
3. Combine the masses of the reactants:
- Since the masses directly add up in a 1:1 reaction, we simply sum the masses of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] to find the total mass of [tex]\( AB \)[/tex].
4. Sum of the masses:
- The mass of [tex]\( A \)[/tex] is 10 grams.
- The mass of [tex]\( B \)[/tex] is 45 grams.
- Therefore, the total mass of [tex]\( AB \)[/tex] = 10 grams + 45 grams = 55 grams.
Thus, Colin would make 55 grams of substance [tex]\( AB \)[/tex]. Therefore, the correct answer is [tex]\( \boxed{55 \, \text{g}} \)[/tex].
1. Identify the given quantities of the reactants:
- Colin added 10 grams of substance [tex]\( A \)[/tex].
- Colin also added 45 grams of substance [tex]\( B \)[/tex].
2. Understand the reaction:
- The reaction is given by [tex]\( A + B \rightarrow AB \)[/tex], indicating a 1:1 molar ratio. This implies that one mole of [tex]\( A \)[/tex] reacts with one mole of [tex]\( B \)[/tex] to form one mole of [tex]\( AB \)[/tex].
3. Combine the masses of the reactants:
- Since the masses directly add up in a 1:1 reaction, we simply sum the masses of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] to find the total mass of [tex]\( AB \)[/tex].
4. Sum of the masses:
- The mass of [tex]\( A \)[/tex] is 10 grams.
- The mass of [tex]\( B \)[/tex] is 45 grams.
- Therefore, the total mass of [tex]\( AB \)[/tex] = 10 grams + 45 grams = 55 grams.
Thus, Colin would make 55 grams of substance [tex]\( AB \)[/tex]. Therefore, the correct answer is [tex]\( \boxed{55 \, \text{g}} \)[/tex].