Answer:
The balanced equation is:
2 HNO3 + Mg ---> Mg(NO3)2 + H2
From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg
Moles of Mg:
Molar mass of Mg = 24 g/mol
Moles = Given mass / Molar Mass
Moles of Mg = 4.47 / 24 = 0.18 moles (approx)
Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed
Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3
Unreacted moles of HNO3 = Total Moles - Moles consumed
Unreacted moles of HNO3 = 0.64 moles (approx)
Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate
From the given options, we can see that 0.632 moles is the closest value to our answer
Therefore, 0.632 moles will remain after the reaction
<h3><u>Answer;</u></h3>
10.80 ° C
<h3><u>Explanation;</u></h3>
From the information given;
Initial temperature of water = 24.85°C
Final temperature of water = 35.65°C
Mass of water = 1000 g
The specific heat of water ,c = 4.184 J/g °C.
The heat capacity of the calorimeter = 695 J/ °C
Change in temperature ΔT = 35.65°C - 24.85°C
= 10.80°C
Answer:
V=0.68L
Explanation:
For this question we can use
V1/T1 = V2/T2
where
V1 (initial volume )= 0.75 L
T1 (initial temperature in Kelvin)= 303.15
V2( final volume)= ?
T2 (final temperature in Kelvin)= 273.15
Now we must rearrange the equation to make V2 the subject
V2= (V1/T1) ×T2
V2=(0.75/303.15) ×273.15
V2=0.67577931717
V2= 0.68L
