A fatty acid with 10 carbons will be absorbed via the
A sample of gas has a volume of 5.58 l at a pressure of 715 mm hg. what is the volume of the gas when the pressure of the gas is increased to 755 mm hg?
Balance the reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coefficient is "1." Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃
Answer:
blank, 2, blank, 2
The process that changes a solid directly into a gas is called sublimation. A solid that begins with a mass of 450 g will sublime to give off a gas with a mass of _______ g.
Answer:
The answer is 450 g
Explanation:
When the solid sublimes, it gives the same mass of gas because sublimation is a physical change, and it is valid the energy and mass conservation law (mass cannot be destroyed or created in a chemical or physical process). If sublimation is complete, all the molecules changes its physical state from solid to gas.
Sublimation is the process where a solid changes directly into a gas without passing through the liquid phase. The mass of the substance remains unchanged, so a solid with a mass of 450 g will yield a gas with a mass of 450 g.
The process that changes a solid directly into a gas is called sublimation. In this process, a solid transitions to a gas phase without passing through the liquid phase. Importantly, the mass of the substance remains constant during this phase change. Therefore, if you start with a solid that has a mass of 450 g, the resulting gas will also have a mass of 450 g. This is in accordance with the law of conservation of mass which suggests that mass can neither be created nor be destroyed.
An example of sublimation occurs with dry ice (solid CO₂), which transitions directly from a solid to a gas at standard temperature and pressure.
What is the half-life (in seconds) of a zero-order reaction which has an initial reactant concentration of 0.884 M with a k value of 5.42 × 10–2 M/s?
Answer:
The half-life (in seconds) of a zero-order reaction is 8.15 seconds.
Explanation:
Initial concentration of the of the reactant = [tex][A_o]=0.884 M[/tex]
The value of rate constant = [tex]k=5.42\times 10^{-2} M/s[/tex]
The half life for zero order reaction is given as:
[tex]t-{\frac{1}{2}}=\frac{[A_o]}{2k}=\frac{0.884 M}{2\times 5.42\times 10^{-2} M/s}=8.15 s[/tex]
The half-life (in seconds) of a zero-order reaction is 8.15 seconds.
What is the structural formula of 2-methylbutan-2-ol (sometimes called 2-methyl-2-butanol)?
The rate constant for a second-order reaction is 0.54 m-1s-1. what is the half-life of this reaction if the initial concentration is 0.30 m?
The correct answer is: 6.173 seconds.
Explanation:
The formula for half life (of reaction) is:
Half-life = [tex]\frac{1}{K*I}[/tex] --- (A)
Where,
K = Rate constant = 0.54 [tex]\frac{1}{ms}[/tex]
I = Initial concentration = 0.30 m
Plug in the values in equation (A):
Half-life = [tex]\frac{1}{0.54*0.30} = 6.173[/tex]
Hence, the half life is 6.173 seconds.
The half-life of a second-order reaction is [tex]\boxed{{\text{6}}{\text{.173 seconds}}}[/tex]
Further explanation:
Second-order reaction:
A reaction is said to be of second-order if its rate is proportional to the square of the concentration of one reactant. The general example of second-order reaction is,
[tex]2{\text{A}}\to{\text{P}}[/tex]
Here,
A is the reactant.
P is the product.
The rate is calculated by using the following equation:
[tex]{\text{Rate}}={\text{k}}{\left[{\text{A}}\right]^2}[/tex]
Another form of second-order reaction is as follows:
[tex]{\text{A}}+{\text{B}}\to{\text{P}}[/tex]
Here,
A and B are the two different reactants.
P is the product.
The rate is calculated by using the following equation:
[tex]{\text{Rate}}={\text{k}}\left[{\text{A}}\right]\left[{\text{B}}\right][/tex]
Half-life:
Half-life is defined as the time required to reduce the concentration of reactant to half of its initial value. It is denoted by [tex]{{\text{t}}_{1/2}}[/tex] . The general expression to calculate [tex]{{\text{t}}_{1/2}}[/tex] of second-order reaction is,
[tex]{{\text{t}}_{1/2}}=\frac{1}{{{\text{k}}{{\left[{\text{A}}\right]}_{\text{0}}}}}[/tex] …… (1)
Here,
[tex]{{\text{t}}_{1/2}}[/tex] is the half-life of the reaction.
k is the rate constant for the reaction.
[tex]{\left[ {\text{A}}\right]_{\text{0}}}[/tex] is the initial concentration of the reactant.
The rate constant for the given reaction is [tex]0.54\;{{\text{M}}^{-1}}{{\text{s}}^{-1}}[/tex] .
The initial concentration of the given reaction is 0.30 M.
Substitute these values in equation (1).
[tex]\begin{aligned}{{\text{t}}_{1/2}}&=\frac{1}{{\left({0.54\;{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}}\right)\left( {{\text{0}}{\text{.30 M}}}\right)}}\\&=\frac{1}{{0.162\;{{\text{s}}^{ - 1}}}}\\&=6.1728\;{\text{s}}\\&\approx 6.173\;{\text{s}}\\\end{aligned}[/tex]
So, the half-life of a second-order reaction is 6.173 seconds.
Learn more:
1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
2. Calculate the moles of ions in the solution: https://brainly.com/question/5950133
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Chemical Kinetics
Keywords: second-order reaction, half-life, initial, half, 6.173 seconds, 0.30 m, k, A0, A, B, P, 2A, reactant, product, t1/2.