Answer:
d. there are no conditions for this correlation.
Explanation: Dispersal of matter is a term used to describe the various processes and activities involved in the transfer of materials from one location to another.
The spontaneity of a process tries to explain that a process will continue to occur when an initial force has been applied, it doesn't require the continous input of energy for it to continue to take place.
Answer: 160 ml
Explanation:
The expression used will be :
where,
= concentration of Ist alcohol solution= 10%
= concentration of 2nd alcohol solution= 55%
= volume of Ist alcohol solution = 200 ml
= volume of 2nd alcohol solution= v ml
= concentration of resulting alcohol solution= 30%
= volume of resulting alcohol solution= (v+200) ml
Now put all the given values in the above law, we get the volume of added.
By solving the terms, we get :
Therefore, the volume of 55% mixture needed to be added to obtain the desired solution is 160 ml.
Answer:
Explanation:
Our
sample yielded 1g of
and 16g of
, but our unknown sample yielded 2 times as much
for the same amount of
.
What does this mean? that the H:O proportion for the unknown sample is twice the H:O proportion for the
sample.
What is the H:O proportion for the
sample? As we can see from its formula, it's 1:1, therefore the proportion for the unknown formula must be 2:1.
That means, two H atoms for every O atom. We could write that as:
and you should recognize that formula, for it is one of the most common compounds on earth, Water.
26. Given the nuclear equation: 58 scu Ni + X 58 What nuclear particle is represented by X? (1) - 26251699.
Answer:
Heat released =811.68kcal
Explanation:
2 CO(g) + O2(g) → 2 CO2(g) ΔH=-135.28kcal
for this type of question 1st balance the chemical reaction and use unitry method.
From the above balanced equaion it is clearly that,
2 mole of CO reacts with 1 mole of O2 and we have sufficient amount of oxygen means excess amount.
and also for complete consumption of 2 mole of CO, ΔH=-135.28 kcal
for complete consumption of 12 moles of CO2 ΔH=
Heat released =811.68kcal