Volume Ba(OH)2 = 23.4 mL in liters :
23.4 / 1000 => 0.0234 L
Molarity Ba(OH)2 = 0.65 M
Volume HNO3 = 42.5 mL in liters:
42.5 / 1000 => 0.0425 L
number of moles Ba(OH)2 :
n = M x V
n = 0.65 x 0.0234
n = 0.01521 moles of Ba(OH)2
Mole ratio :
<span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
</span>
1 mole Ba(OH)2 ---------------- 2 moles HNO3
0.01521 moles ----------------- moles HNO3
moles HNO3 = 0.01521 x 2 / 1
moles HNO3 = 0.03042 / 1
= 0.03042 moles HNO3
Therefore:
M ( HNO3 ) = n / volume ( HNO3 )
M ( HNO3 ) = 0.03042 / 0.0425
M ( HNO3 ) = 0.715 M
Non metals
(It’s both halogens and noble gases ) the other ones which are white(not highlighted) are alkali metals, alkaline earth metals, and transition metals :3
Answer:
the initial temperature of the iron sample is Ti = 90,36 °C
Explanation:
Assuming the calorimeter has no heat loss to the surroundings:
Q w + Q iron = 0
Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )
Assuming Q= m*c*( T- Tir)
mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0
Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )
Tir = 90.36 °C
Note :
- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C
- We assume no reaction between iron and water
Li2CO3 + 2HNO3 ----> 2LiNO3 + H2O + CO2
Iron filings can be attracted by the magnet whereas salt can not. So the mixture can be separated by a magnet
