Name any two metals that are found in liquid state at NTP.

Chemistry

1 answer:

garri49 [273]2 years ago

4 0

Answer:

mercury and gallium are two metals that are found in liquid state at NTP.

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A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and

zlopas [31]

Answer:

37.1°C.

Explanation:

Firstly, we need to calculate the amount of heat (Q) released through this reaction:

∵ ΔHsoln = Q/n

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

The negative sign of ΔHsoln indicates that the reaction is exothermic.

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

We can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

∴ final temperature = 25°C + 12.1 = 37.1°C.

6 0

2 years ago

When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr

konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.