The main idea of an essay is usually portrayed in the thesis.
Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
Q water = m * C * (Tf - Ti)
= (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol
Answer:
3,29L
Explanation:
3.29L = V2
Formula: V1/T1 = V2/T2
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Given:
V1 = 3.0 L V2 = ?
T1 = 310 K T2 = 340 K
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Plugin:
(X stands in place of V2 just to make it easier to look at)
[3.0L / 310K = X / 340K]
(3.0L / 310K = 0.01L/K)
0.01L/K = X / 340K
(multiply 340K on both sides, it cancels out on the right)
0.01L/K * 340K = X
(0.01L/K * 340K = 3.29L)
**3.29L = X**
[or]
**3.29L = V2**
Answer:
Strong acid
Explanation:
An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
Hydroxonium ions are formed as a result of the chemical bonding between the oxygen of water molecules and the protons released by the acid due to its ionisation. This makes aqueous solution of acids conduct electricity.
A strong acid is one that ionizes almost completely. Examples are:
1. Hydrochloric acid
2. Tetraoxosulphate (VI) acid
3. Trioxonitrate (V) acid
4. Hydroiodic acid
5. Hydrobromic acid
