The final temperature is -138 °C.
Explanation:
Using the equation of specific heat
We can easily find the final temperature of a 73.174 g of copper sample. As we know that specific heat is the amount of energy required to raise the temperature of the object to 1°C.
The specific heat of copper is known as 0.387 J/g°C and the initial temperature is said as 102 °C . The mass is given as 73.174 g. The heat released is 6800 J.
Since the heat is released the Q value will be negative.
Thus, the final temperature is -138 °C.
Answer:
See explanation
Explanation:
2HCl(aq) + CaCO3(aq) ------->CaCl2(aq) + CO2(g) + H2O(l)
Number of moles of acid present = 50/1000 * 0.15 = 0.0075 moles
Number of moles of calcium carbonate = 0.054g/100 g/mol = 0.00054 moles
2 moles of HCl reacts with 1 mole of calcium carbonate
x moles of HCl reacts with 0.00054 moles of calcium carbonate
x = 2 * 0.00054/1
x = 0.00108 moles of HCl
Amount of acid left = 0.0075 moles - 0.0075 moles = 0.00642 moles
Reaction of HCl and NaOH
HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)
Since the reaction is in the mole ratio of 1:1
0.00642 moles of HCl is neutralized by 0.00642 moles of NaOH
The balanced chemical reaction would be:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:
Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl
0.646 - 0.48 = 0.166 mol HCl left
Answer:
660kcal
Explanation:
The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.
We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:
3.3L * (1000mL / L) * (5g/100mL)= 165 g.
If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.
