Answer:
239.7 g
Explanation:
Step 1: Write the balanced equation
2 LiBr + I₂ → 2 LiI + Br₂
Step 2: Convert the molecules of iodine to moles
We have 9.033 × 10²³ particles (molecules) of iodine. In order to convert molecules to moles, we will use the <em>Avogadro's number</em>: there are 6.022 × 10²³ molecules of iodine in 1 mole of iodine.
Step 3: Calculate the moles of bromine produced
The <em>molar ratio of I₂ to Br₂</em> is 1:1. Then, the moles of bromine produced are 1.500 moles.
Step 4: Calculate the mass of bromine
The <em>molar mass of bromine</em> is 159.81 g/mol. The mass corresponding to 1.500 moles is:
Answer: negative acellaration or mass.
Explanation:
the first reason why is that i got that quistion right. and when objects are unbalanced it gives negative acellaration
Answer:
The boiling point of sample X and sample Y are exactly the same.
Explanation:
The difference between sample X and sample Y is that they occupy different volumes. However, they both contain pure water. Remember that pure water has uniform composition irrespective of its volume.
Volume does not affect the boiling point as long as the volume is small enough not to give rise to significant pressure changes in the liquid.
The boiling point of a liquid is the temperature at which the pressure exerted by the surroundings upon a liquid is equaled by the pressure exerted by the vapour of the liquid; under this condition, addition of heat results in the transformation of the liquid into its vapour without raising the temperature.
It can be clearly seen from the above that the volume of a solution of pure water does not affect its boiling point hence sample X and sample Y will have the same boiling point.
Answer: Aqua fortis is also known as Nitric Acid
Explanation: and the process in making aqua fortis is THE OSTWALDS PROCESS
detailed explanation by me
ammonia is also used in this process, which is reacted with oxygen and water and a catalyst platinum.
HOPE THIS HELPSS!!!
Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
