Answer:
2 .Sulphurous acid
Explanation:
Sulphur dioxide can dissolve in water to form Sulfurous acid(H2SO3). sulphurous acid is weackly dibasic acid. sulphur dioxide is a major component of acid rain since it mixes with vapour in the atmosphere reacting to produce H2So4 .
Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
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.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
The the exact mass is 24.740% (according to the internet) but I’d say that it’s 25
Glucose is extracted in the urine and all of the glucose is reabsorbed.
Answer:
The answer is 3-Phenylpropanoic acid (see attached structure)
Explanation:
From spectral data:
3005 cm-1 ⇒ carboxylic acid (broad band)
1670 cm-1 ⇒ C=C
1603 cm-1 ⇒ Aromatic C-C bond
H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding (clearer investigation of spectrum would be expedient).
Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.
