Answer: 72.41% and 26.90% respectively.
Explanation:
At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.
We can calculate recovery as:
So the answer to the first question is 72.41%.
For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.
So the answer to the second question is 26.90%.
Answer:
37S
Explanation:
Radioactivity is the spontaneous emission of particles and / or electromagnetic radiation by unstable atomic nuclei leading to their disintegration.
We have two main types of radioactivity: radioactive decay and artificial transmutation.
In radioactive decay ( natural radioactivity ), a naturally occurring radioactive element like Uranium-238 disintegrates or decays into more stable isotopes with the emission of particles and/or radiation.
23892U = 23490Th + 42He
Artificial transmutation is the collision of two particles where one particle captures the other used to bombard it. There is subsequent production of isotopes similar or different from the bombarded particle. Neutrons, alpha particles ( helium nucleus ), electrons, protons can be used to bombard elements.
147N + 42He = 178O + 11P
For the above question which is artificial transmutation, the reaction equation is
4018Ar + 10n = 3716S + 42He
So, the neutron capture by Argon-40 will produce a radioisotope Sulphur-37 with the emission of an alpha particle.
Answer:
Option 2= Glucose
Explanation:
Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.
Facilitated diffusion:
it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.
Primary active transport:
The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.
Secondary active transport:
It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.
GaBr3
Gallium=Ga
Bromine= Br
Bromide=Br3
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Correct Answer:
Option 3 i.e. 30 g of KI dissolved in 100 g of water.
Reason:
Depression in freezing point is a
colligative property and it is directly proportional to molality of solution.
Molality of solution is mathematically expressed as,
Molality = </span>
<span>
In case of
option 1 and 2, molality of solution is
0.602 m. For
option 3, molality of solution is
1.807 m, while in case of
option 4, molality of solution is
1.205 m.
<u><em>Thus, second solution (option 2) has highest concentration (in terms of molality). Hence, it will have lowest freezing point</em></u></span>
