Thermal energy will flow from an object high temperature to an object of low one. In this case, the thermal energy will flow from object B to object A.
32.73 g phosphorus
42.27 g oxygen.
= 74 .99 substance
The balanced chemical equation is represented as:
<span>Ba(OH)2·8 H2O + 2 NH4SCN = Ba(SCN)2 + 10 H2O + 2 NH3
To determine the mass of </span>ammonium thiocyanate needed completely react, we use the amount given of the other reactant. Then, convert to moles by the molar mass. And using the relations of the substances in the reaction, we can determine the amount of <span>ammonium thiocyanate.
moles </span>Ba(OH)2·8 H2O = 5.7 g Ba(OH)2·8 H2O ( 1 mol / 315.46 g ) = 0.0181 mol Ba(OH)2·8 H2O
moles NH4SCN = 0.0181 mol Ba(OH)2·8 H2O ( 2 mol NH4SCN / 1 mol Ba(OH)2·8 H2O ) = 0.0361 mol NH4SCN
mass NH4SCN = 0.0361 mol NH4SCN ( 76.122 g / 1 mol ) = 2.75 g NH4SCN
Explanation:
It is given that energy to transfer one water molecule is 
J/molecule
As it is known that in 1 mole there are 
atoms.
So, energy in 1 mole = 
J/mol
= 13.3 kJ/mol
As, 
Putting the given values in the above formula as follows.
= 0.08377
= 1.213 = 
= 
= 
= 
water molecules per oil molecule
Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is .
Answer:
333.7 g.
Explanation:
- The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>
Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).
Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>
molaity (m) is the no. of moles of solute per kg of the solvent.
∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))
∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.
∵ no. of moles = mass/molar mass.
<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>
