The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:
-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)
where C is the specific heat capacities of the materials.
We calculate as follows:
-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 -----> OPTION C
830 mL. A 2.3 mol/L solution of CaCl2 has a volume of 830 mL
I am guessing that the concentration of your solution is 2.3 mol/L.
a) Moles of CaCl2
MM of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)
= 1.910 mol CaCl2
b) Volume of solution
V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution
= 830 mL solution
Answer: try to understand coz the question is not valid
Explanation: Explain the relationship between forward and reverse reactions at equilibrium and predict how changing the amount of a reactant or product (creating a stress) will affect that relationship.For example (select one from each underlined section)If the amount of (reactant or product) increases, the rate of the (forward or reverse)reaction will (increase or decrease)to reach a new equilibrium. If the amount of (reactant or product) decreases, the rate of the (forward or reverse)reaction will (increase or decrease)to reach a new equilibrium. Procedure: Access the virtual lab and complete the inquiry experiment
I believe it is the Great Plains in Nebraska.
Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:
a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:
Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:
c) The leftover is computed as follows:
NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.
