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Helen [10]
3 years ago
14

Synthetic, or man-made, elements typically exist for only a short time

Chemistry
1 answer:
o-na [289]3 years ago
8 0
This is true. Elements past lead are radioactive, because the repulsive force of the protons cannot be overpowered by the “gluing” ability of neutrons (remember, likes repel). As more and more protons are added, generally, the elements become more unstable; for example, Bismuth, right next to lead on the Periodic Table, is radioactive, but the half life of this element is about a billion times longer than the current age of the universe, but Oganesson, element number 118, has a half life of fractions of a second.
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The original gas volume in a balloon is 50.mL when the pressure is 1.00 atm. Predict the new gas volume when the pressure is 3 t
damaskus [11]
According to Boyle's law, Volume is indirectly proportional to the pressure.

So, P1V1 = P2V2

Substitute the known values, 
1 * 50 = 3 * V2

V2 = 50/3 = 16.67 mL

In short, Your Answer would be: 16.67 mL

Hope this helps!
5 0
3 years ago
Question 1
Maksim231197 [3]

Answer:

Hydrogen bonding, Dipole-dipole

Explanation:

Let us look closely at the structure of H2S, this will enable us to decide if it will show dipole-dipole interaction or not. Given that Sulphur has a greater electro negativity compared to Hydrogen, we expect that the S – H bond will be a polar bond with the direction of polarity pointing towards the sulphur atom according to the usual convention in chemistry. Now, we know that H2S is a bent molecule (there are two lone pairs on the central sulphur atom which leads to a bent molecular geometry), therefore, the vectorial sum of the bond dipole moments will produce a non- zero total dipole moment. This implies that there is a permanent non-zero dipole moment in H2S therefore we expect the molecule to exhibit dipole-dipole interactions .

Similarly, we know that hydrogen bonding exists when hydrogen is bonded to a highly electronegative element such as fluorine, sulphur, oxygen etc. In H2S, hydrogen is bonded to sulphur hence we expect that there should be hydrogen bonding between H2S molecules. Hydrogen bonding is actually a type of dipole-dipole interaction.

6 0
3 years ago
C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????
Yanka [14]

Answer:

Rate constant at 725 K is 5.2\times 10^{-4}s^{-1}

Explanation:

According to Arrhenius equation for a reaction-

ln(\frac{k_{2}}{k_{1}})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})

where k_{2} and k_{1} are rate constants of reaction at T_{2} and T_{1} temperatures (in kelvin) respectively.

E_{a} is activation energy of reaction.

Here T_{1}= 600 K , k_{1}= 6.1\times 10^{-8}s^{-1}

T_{2}= 725 K, E_{a}= 262 kJ/mol and R = 8.314 J/(mol.K)

So plugin all the values in the above equation-

ln(\frac{k_{2}}{6.1\times 10^{-8}s^{-1}})=\frac{262\times 10^{3}J/mol}{8.314J/(mol.K)}\times (\frac{1}{600K}-\frac{1}{725K})

So, k_{2} = 5.2\times 10^{-4}s^{-1}

Hence rate constant at 725 K is 5.2\times 10^{-4}s^{-1}

7 0
3 years ago
Calculate the freezing point of an aqueous 0.10 m fecl3 solution using a van't hoff factor of 3.2.
Black_prince [1.1K]

ccccccccccccccccccccccccccccccccccc

6 0
3 years ago
2. You are making jelly. After mixing the ingredients you pour the liquid in a bowl, co
Andrei [34K]

b.

Same because, mass doesn’t count liquid or solid

3 0
3 years ago
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