The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C
<h3>How to calculate temperature?</h3>
The initial temperature of the copper metal can be calculated using the following formula on calorimetry:
Q = mc∆T
mc∆T (water) = - mc∆T (metal)
Where;
- m = mass
- c = specific heat capacity
- ∆T = change in temperature
According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:
400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)
30,096 = 93.6T - 2246.4
93.6T = 32342.4
T = 345.5°C
Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.
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Answer:
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Explanation:
Aluminum foil does contain billions of atoms, but we write them just as Al because aluminum does not exist in single atom forms in normal circumstances but as billions in metallic bonds.
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Answer:
ACTIVITY 1
Sample 1 has a stronger taste of lemon, and is more sour.
Sample 2 has a sweeter taste, my guess is because there's more sugar:lemon juice ratio.
0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
<h3>What is dilution?</h3>
Dilution is a process of making a solution of lower concentration from a solution of higher concentration by the addition of solvent to a given volume of the solution of higher concentration.
Dilution of solutions is done using the dilution formula in order to determine the given volume of diluent or stock solution required. The dilution formula is given below:
where:
- C1 = Initial concentration of enzyme
- C2 = Final concentration of enzyme
- V1 = Initial volume
- V2 = Final volume
For the enzyme dilution;
C1 = 1 mg/mL
C2 = 1/50 mg/ml = 0.02 mg/ml
V= ?
V2 = 5 ml
V1 = C2V2/C1
V1 = 0.02 * 5/1 = 0.1 mL
Therefore, 0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
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