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3 years ago

PLEASE HELP ME!!!!!!!!

Chemistry

2 answers:

ch4aika [34]3 years ago

7 0

So it has to be either the second or the third one, so let's focus on that. If the substance can make energy in a power plant, then it is most likely Uranium. Uranium is usually radioactive. Now the third one talks about fission, where the atoms that are unstable break apart into smaller atoms to become more stable. I'd say the second one, because it is obviously radioactive.

Mazyrski [523]3 years ago

6 0

I think the answer is c

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Please help I need to correct I don't understand it

victus00 [196]

Answer is d. in hetrogeneous you can separate things from each other

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2 years ago

The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime

krok68 [10]

Answer: The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = $2.83\times 10^{-3}$

a - x = amount left after decay process = $3.06\times 10^{-4}$

$559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=3.96\times 10^{-3}min^{-1}$

The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

6 0

3 years ago

Controlled nuclear chain reactions

Lunna [17]

The nuclear reactions which are under experimenter's control are said to be controlled nuclear reactions. In this, you can maintain the speed of the incident particle. α and β-decay process are examples of non-controlled nuclear reactions.

8 0

3 years ago

A certain sample of a liquid has a mass of 42 grams and a volume of 35 centimeters3. What is the density of the liquid? The dens

Xelga [282]

Density of liquid= $\frac{mass of liquid}{volume of liquid}$ .

so, density of liquid= $\frac{42 gm}{35 cm^{3} }$ = 1.2 gm/cm³.

8 0

2 years ago

Read 2 more answers

Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of

natima [27]

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several energy values for each interaction present in the Newman structures:

-) Methyl-methyl gauche: 3.8 KJ/mol

-) Methyl-H eclipse: 6.0 KJ/mol

-) Methyl-methyl eclipse: 11.0 KJ/mol

-) H-H eclipse: 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

Molecule A

In this molecule, we have 2 Methyl-methyl gauche interactions only, so:

(3.8x2) = 7.6 KJ/mol

Molecule B

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

Molecule C

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule D

In this molecule, we have three Methyl-H eclipse interaction, so:

(6*3) = 18 KJ/mol

Molecule E

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule F

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

3 0

3 years ago