Answer:
3626.76dm³
Explanation:
Given parameters:
Number of moles of Nitrogen in tank = 17moles
Temperature of the gas = 34°C
Pressure on the gas = 12000Pa
Unkown:
Volume of the tank, V =?
Converting the parameters to workable units:
We take the temperature from °C to Kelvin
K = 273 + °C = 273 + 34 = 307k
Taking the pressure in Pa to atm:
101325Pa = 1atm
12000Pa = 0.118atm
Solution:
To solve this problem, we employ the use of the ideal gas equation. The ideal gas law combines three gas laws which are the Boyle's law, Charles's law and the Avogadro's law.
It is expressed as PV = nRT
The unknown is the Volume and we make it the subject of the formula
V = 
Where R is called the gas constant and it is given as 0.082atmdm³mol⁻¹K⁻¹
Therefore V = 
= 3626.76dm³
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
Sodium borohydride is a relatively selective reducing agent Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.
The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
to learn more about sodium borohydride and ethanol click here ,
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Döbereiner grouped the known elements into <em>triads</em> (sets of three) so that
• The <em>atomic mass of the middle element</em> was approximately the average of the other two
• The <em>chemical properties of the middle element</em> were between those of the other two
• The <em>physical properties of the middle element</em> were between those of the other two
One example of a triad is Li – Na – K.
(a) Atomic mass of Na = 23.0 u
Average atomic mass of Li and K = (6.9 u + 39.1 u)/2 = 46.0 u/2 = 23.0 u
(b) Li reacts slowly with water. Na reacts rapidly. Potassium reacts violently.
(c) Melting point of Na = 371 °C.
Average melting point of Li and K = (454 °C + 330 °C)/2 = 784 °C/2
= 392 °C
1: True
2: True
3: False
4: False
(Question 2 might not be true, not sure)
