Answer:
ΔH° = 840 kJ/mol
Explanation:
Let's consider the following balanced reaction:
2 AgNO₃(aq) + CaCl₂(aq) ⇄ 2 AgCl(s) + Ca(NO₃)₂
Then, we need to know the moles of both reactants:
AgNO₃: n = 0.200 mol/L × 0.0500 L = 0.0100 mol
CaCl₂: n = 0.100 mol/L × 0.0500 L = 0.00500 mol
According to the balanced equation we need 2 moles of AgNO₃ per each mole of CaCl₂, and this coincides with the experimental data, so there is no limiting reactant. Let's use AgNO₃ to find out how many moles of AgCl are produced.
Now, we can calculate the total amount of heat released using the following expression:
Q = c × m × ΔT
where,
c is the heat capacity of the solution
m is the mass of the solution
ΔT is the change in temperature (26.0 °C - 25.0°C = 1.00 °C)
Since the volume is 100.0 mL (50.0 mL + 50.0 mL) and the density is 1.05 g/mL, we can calculate the mass of the solution like:
m = 1.05 g/mL × 100.0 mL = 105 g
Then,
Finally,
Answer:
violet
Red has the longest wavelength and violet has the shortest wavelength. When all the waves are seen together, they make white light. Ultraviolet (UV) light—is radiation with a wavelength shorter than that of visible light, but longer than X-rays, in the range 10 nm to 400 .
Explanation:
0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.
If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?
A. less than 20∘C
B. 20∘C
C. greater than 20∘C
Answer:
Option C is correct
Explanation:
Increase in the mass of aluminium block would increase the heat capacity of block isothermaly before immersed in water at 0° so heat available for transfer is higher so equilibrium temperature of system would increase.
Answer:
1. kmol of methanol= 3.12 Kmol
2. kmol of water= 5.55 Kmol
3. Liters of methanol= 126.4 L
4. L of water= 100 L
Explanation:
1. kmol of methanol?
32.04 kg methanol ______________ 1 kmol of methanol
100 kg of methanol_______________ X= 3.12 kmol ofmethanol
2. kmol of water?
18.01 kg water ______________ 1 kmol of wáter
100 kg of wáter_______________ X= 5.55 kmol of water
3. Liters of methanol?
0.791 kg methanol _______________________1.00 L of methanol
100kg methanol _______________________x= 126.4 L of methanol
4. L of water?
1kg water _______________________1.00 L of water
100kg water _______________________x= 100 L of water
The theoretical yield of a reaction is 28% if 25.0 grams of product were actually produced from a reaction that has a 88% yield .
The theoretical yield of a reaction is calculated in following steps .
Identify the limiting reagent, which is the reagent with the fewest moles.
Divide the fewest number of reagent moles by the stoichiometry of the product.
Multiply the result of Step 3 by the molecular weight of the desired product.
theoretical yield : the maximum amount of product that can be produced from a given amount of reactant. this quantity of product is rarely produced during a chemical reaction. limiting reactant.
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