Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
To begin with, the equation given is not correct.
Correct equation is : CaCO3 + HCl ---> CaCl2 + H2O + CO2
It's CaCl2 not CaCl because Ca has a valency of 2
LHS RHS
CaCO3 + HCl ---> CaCl2 + H2O + CO2
First of all, to balance the equation you must look at the number of atoms on each side of the equation.
we have 2 H on the RHS and 1 H on the LHS. So, we put a 2 on the LHS
CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Check for the LHS: 1 Ca, 1 C, 3 O, 2 H & 2 Cl on the LHS
Now check for the RHS: 1 Ca, 2 Cl, 2 H, 1 C & 3 O
Hope it helped!
Answer:
1/360
Explanation:
let x = liters
molarity=moles of solute/liters of solution, 7.2=0.02/x or 7.2=(1/50)(1/x), 7.2(50)=(1/x), 360(x)=1, x=1/360
Molar mass :
NaBr = 103 g/mol
Pb(NO3)2 = 331.20 g/mol
<span><span /><span>Balanced chemical equation :
</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2
331.20 g = 2*103*311
331.20 g = 64066
mass ( NaBr ) = 64066 / 331.20
mass ( naBr) = 193,43 g of NaBr
hope this helps!.
</span>
