The specific heat of a substance is heat released or absorbed by one gram for a change in 1 degree celsius
so higher the specific heat, more the heat required to raise the temperature by same extent or more the heat released on decrease in temperature upto same extent (for two substance)
Here the specific heat of sandy clay is 1.4 J/g °C
So it is less than that of wet mud hence sandy clay is needs to release less energy to decrease its temperature
Answer: (3) 15
Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.
Answer:
The average kinetic energy of the gas particles is greater in container B because it has a higher temperature.
Explanation:
<em>The correct option would be that the average kinetic energy of the gas particles is greater in container B because it has a higher temperature.</em>
<u>According to the kinetic theory of matter, the temperate of a substance is a measure of the average kinetic energy of the molecules of substance. In other words, the higher the temperature of a substance, the higher the average kinetic energy of the molecules of the substance.</u>
In the illustration, the gas in container B showed a higher temperature than that of container A as indicated on the thermometer, it thus means that the average kinetic energy of the molecules of gas B is higher than those of gas A.
Answer:
by adding water into the mix
Explanation:
this will dissolve the salt
Answer:
0.70 g
41 %
Explanation:
We can write the Williamson ether synthesis in a general form as:
R-OH + R´-Br ⇒ R-O-R´
where R-OH is an alcohol and R´-Br is an alkyl bromide.
We then see that the reaction occurs in a 1:1 mole ratio to produce 1 mol product.
Therefore what we need to calculate the theoretical yield and percent yield is to compute the theoretical number of moles of 2-butoxynaphthalene produced from 0.51 g 2-naphthol, and from there we can calculate the percent yield.
molar mass 2-naphthol = 144.17 g/mol
moles 2-naphthol = 0.51 g / 144.17 g/mol = 0.0035 mol 2-naphthol
The number of moles of produced:
= 0.0035 mol 2-naphthol x ( 1 mol 2-butoxynaphthalene /mol 2-naphthol )
= 0.0035 mol 2-butoxynaphthalene
The theoretical yield will be
= 0.0035 mol 2-butoxynaphthalene x molar mass 2-butoxynaphthalene
= 0.0035 mol x 200.28 g/ mol = 0.70 g
percent yield= ( 0.29 g / 0.70 ) g x 100 = 41 %
