Answer:
1. V2.
2. 299K.
3. 451K
4. 0.25 x 451 = V2 x 299
Explanation:
1. The data obtained from the question include:
Initial volume (V1) = 0.25mL
Initial temperature (T1) = 26°C
Final temperature (T2) = 178°C
Final volume (V2) =.?
2. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K
3. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Final temperature (T2) = 178°C
Final temperature (T1) = 178°C + 273 = 451K
4. Initial volume (V1) = 0.25mL
Initial temperature (T1) = 299K
Final temperature (T2) = 451K
Final volume (V2) =.?
V1 x T2 = V2 x T1
0.25 x 451 = V2 x 299
Answer:
Explanation:
Hello!
In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:
Then, the liters of solution:
Finally, the molarity turns out:
Best regards!
Phosphorus can be prepared from calcium phosphate by the following reaction:
Phosphorite is a mineral that contains 
plus other non-phosphorus-containing compounds. What is the maximum amount of 
that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75% by mass? Assume an excess of the other reactants.
Answer: Thus the maximum amount of that can be produced is 0.345 kg
Explanation:
Given mass of phosphorite = 2.3 kg
As given percentage of phosphorite is = 
According to stoichiometry:
2 moles of phosphorite gives = 1 mole of
Thus 5.56 moles of phosphorite give= 
of
Mass of 
Thus the maximum amount of that can be produced is 0.345 kg
