Answer:
a) ethanol is dehydrated by concentrated sulphuric acid to yield ethene
b) Concentrated phosphoric(V) acid, H3PO4, can be used instead of sulphuric acid
Explanation:
Ethene is prepared in the lab by heating ethanol with concentrated sulphuric acid at 170°C. The mixture contains 1:2 volume ratio of the ethanol and concentrated acid. Ethylhydrogentetraoxosulphate VI is first formed and then decomposes to yield ethene and sulphuric acid.
The overall reaction is akin to the dehydration of ethanol. As is easily seen above; this is a two stage reaction. The first stage is the formation of the ester, the second stage involves the dehydration of the ester to yield ethene. The ethene gas is passed through concentrated sodium hydroxide to remove any gaseous impurities present and then it is collected over water.
Concentrated phosphoric(V) acid, H3PO4, can be used instead of concentrated sulphuric acid as a dehydrating agent in the laboratory preparation of ethene.
The best method to solve how many grams of sodium nitrate that will be dissolve in water at 20 degrees Celsius is to use solubility table. Based on the solubility table, 88.3 g will dissolve in water in a 100g NaNO3. Therefore, get the percentage and multiply it to 50g. The answer is 44.15g
The answer is the second choice (B)
EXPLANATION:
In a longitudinal wave the particle displacement is parallel to the direction of wave propagation. The animation at right shows a one-dimensional longitudinal plane wave propagating down a tube. The particles do not move down the tube with the wave; they simply oscillate back and forth about their individual equilibrium positions. Pick a single particle and watch its motion. The wave is seen as the motion of the compressed region (ie, it is a pressure wave), which moves from left to right.
The second animation at right shows the difference between the oscillatory motion of individual particles and the propagation of the wave through the medium. The animation also identifies the regions of compression and rarefaction.
<span>H2SO4 gives 2 moles oh H+ per mole of acid
[H2SO4] = 2M so [H+] = 4M
pH = -log(4) = -.6
Therefore, the pH </span><span>of a 2.0 M H2SO4 solution is -0.6
</span>
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Answer:
8.32 s⁻¹
Explanation:
Given that:
The concentration of myosin = 25 pmol/L
R_max = 208 pmol/L/s
The objective is to determine the turnover number of the enzyme molecule myosin, which has a single active site.
In a single active site of enzyme is known to be a region where there is binding of between substrate molecules, thereafter undergoing chemical reaction.
The turnover number of the enzyme is said to be the number of these substrate molecule which binds together are being converted into products.
The turnover number of the enzyme molecule of myosin can be calculated by the expression: 
⇒ 
= 8.32 s⁻¹
