Explanation:
Upon dissolution of KCl heat is generated and temperature of the solution raises.
Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.
= 4.31 kJ
or, = 4310 J (as 1 kJ = 1000 J)
Mass of solution will be the sum of mass of water and mass of KCl.
Mass of Solution = mass of water + (no. of moles of KCl × molar mass)
= 200 g + 
= 200 g + 13.625 g
= 213.625 g
Relation between heat, mass and change in temperature is as follows.
Q = 
where, C = specific heat of water = 
Therefore, putting the given values into the above formula as follows.
Q =
4310 J = 
Thus, we can conclude that rise in temperature will be 
.
c. a tertiary alcohol; when a ketone reacts with a grignard reagent followed by protonation a tertiary alcohol is formed.
More about tertiary alcohol:
No hydrogen atoms are bonded to the functional group's carbon in a tertiary alcohol. Alcohols that have a hydroxyl group bonded to the carbon atom and are linked to three alkyl groups are referred to as tertiary alcohols. These alcohols' structural makeup largely determines their physical characteristics.
This -OH group's existence enables alcohols to create hydrogen bonds with the atoms next to them. Because of this weak connection, alcohols have higher boiling points than their alkane counterparts.
The alcohol is referred to as a tertiary (3°) alcohol if the carbon atom carrying the alcohol group is connected to three other carbon atoms in the alcohol molecule.
Learn more about tertiary alcohol here:
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Answer:
23.71J is the work that the gas do.
Explanation:
The work that a gas do under isobaric conditions follows the formula:
W = P*ΔV
<em>Where W is work in atmL, P is the pressure and ΔV is final volume -Initial volume In Liters</em>
Replacing with the values of the problem:
W = P*ΔV
W = 0.600atm*(0.44000L - 0.0500L)
W = 0.234atmL
In Joules (1atmL = 101.325J):
0.234atmL × (101.325J / 1 atmL) =
<h3>23.71J is the work that the gas do.</h3>
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Answer: 6 moles
Take a look at the balanced chemical equation for this synthesis reaction
N 2(g] + 3 H 2(g] → 2 NH 3(g]
Notice that you have a 1:3 mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.
So, if you have 2 moles of nitrogen taking part in the reaction, you will need
2 moles N 2 ⋅ 3 moles H 2 /1 mole N 2 = 6 moles H 2
