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enot [183]
3 years ago
11

What is the half-life of A? What will the pressure, initially 32.1 kPa, at

Chemistry
1 answer:
kodGreya [7K]3 years ago
5 0

Answer:

a) 32.09 kPa

b) 32.09 kPa

Explanation:

Given data:

rate constant = 3.56\times 10^{-7} s^{-1}

initial pressure is = 32.1 kPa

half life of A is calculated as

t_{1/2} = \frac{ln 2}{k}

t_{1/2} = \frac{ln 2}{3.56\time 10^{-7}}

t_{1/2} = = 1.956 \times 10^6 s

for calculating pressure we have follwing expression

ln p = ln P_o - kt

P =P_o e^{-kt}

a) P = 32.1 e^{-3.56\times 10^{-7} \times 10} = 32.09 kPa

b) P = 32.1 e^{-3.56\times 10^{-7} \times 10\times 60} = 32.09 kPa

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Arrange the following substances in order of decreasing magnitude of lattice energy. Rank the compounds in order of decreasing m
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The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.

<h3>KCl or NaF, which has a higher lattice energy?</h3>

The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.

<h3>MgO or CaO, which has a larger lattice energy?</h3>

MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.

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3 0
1 year ago
The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins
saul85 [17]
The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>t_{1/2} = 2.5 min

We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{10 min}{2.5 min}
⇒ n=4<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>(1/2)^{n} = x
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3 0
3 years ago
Calculate the Ph and the POH of an aqueous solution that is 0. 040 m in HCl(aq) and 0. 075 m in HBr(aq) at 25 °C.
LenaWriter [7]

pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.

pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.

pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.

Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.

Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :

HCL(0.040M)----> H+(0.040M) +CL-(0.040M)

HBr(0.075M)----> H+(0.075M) +Br-(0.075M)

so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.

pH is calculated as:

pH = -log[H+]

Substituting values in the equation:

log(0.115M)= 0.939 pH

pOH is calculated as:

14 - pH = pOH

Substituting values in the equation above:

14 - 0.939= 13.061 pOH

Therefore, pH is 0.939 and pOH is 13.061.

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5 0
1 year ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
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7 0
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