Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
Potassium sulfide, also
known as dipotassium monosulfide, consists of two potassium ions bonded to a
sulfide atom, rendering the chemical formula K2S.<span>Rarely
found in nature due to its high reactivity with water, potassium sulfide is
refined from the more common potassium sulfate (K2SO4) and is used in many
industries</span>
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
Answer:
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions
Explanation:
We will first write the balanced equation for this scenario
3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2
3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2
The ratio here for both calcium chloride and magnesium nitrate is 
The number of moles of each compound is equal to
Using the mole ratio of 3:2, convert each to moles of sodium phosphate.
mole of CaCl2 is equal to
Na3PO4
mole of CaCl2 is equal to
Na3PO4
Converting moles of sodium phosphate to grams of sodium phosphate we get
g/mol
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions