Answer:
t = 5.7634 s
Explanation:
- A → Pdts
- - rA = K (CA)∧α = - δCA/δt
∴ T = 400°C
∴ α = 1 ....first-order
∴ CAo = 0.950 M
∴ CA = 0.300 M
⇒ t = ?
⇒ - δCA/δt = K*CA
⇒ - ∫δCA/CA = K*∫δt
⇒ Ln (CAo/CA) = K*t
⇒ t = Ln(CAo/CA) / K
⇒ t = (Ln(0.950/0.300)) / (0.200 s-1)
⇒ t = 1.1527 / 0.200 s-1
⇒ t = 5.7634 s
The pressure will increase with decreasing volume. if they remain constant, that is.
Answer:
Explanation:
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In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:
Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.
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When you are collecting DNA, you could be looking for a few different things. A few examples could be skin cells, strands of hair, or possibly even a fingernail. Anything that comes from a person, including blood or saliva can be potential DNA that could help investigators to link a person back to a crime.
Investigators do not need a warrant for analyzing crime scenes due to the fact of the dangers of the fire. You must work quickly because accelerants tend to evaporate within days, sometimes hours. It is also important to note that finding the origin of the fire is very important, to make sure it won't be reignited. Debris is usually cleaned away quickly to ensure health and safety issues.
The point of origin of a fire is the lowest point, since fire burns upwards.
High explosive: Ignite almost instantly, like dynamite and TNT. Two different types are primary and secondary.
<em>Primary: easily ignited, very sensitive to heat and friction. often used to ignite other explosives. </em>
<em>Secondary: much less sensitive to heat and friction, might be ignited using other explosive materials. TNT and dynamite are both secondary. </em>
Low explosive: decompose slowly and include black and smokeless powder. They are the most common type of explosives, and are readily available.
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg
Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol
The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺
The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5
Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol
Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg
