Answer:
Molarity of Sr(OH)₂ = 0.47 M
Explanation:
Given data:
Volume of Sr(OH)₂ = 15.0 mL
Volume of HCl = 38.5 mL (0.0385 L)
Molarity of HCl = 0.350 M
Concentration/Molarity of Sr(OH)₂ = ?
Solution:
Chemical equation:
Sr(OH)₂ + 2HCl → SrCl₂ +2H₂O
Number of moles of HCl:
Molarity = number of moles/ volume in L
0.350 M = number of moles/0.0385 L
Number of moles = 0.350 mol/L× 0.0385 L
Number of moles = 0.0135 mol
Now we will compare the moles of HCl with Sr(OH)₂.
HCl : Sr(OH)₂
2 : 1
0.0135 : 1/2×0.0135 = 0.007 mol
Molarity/concentration of Sr(OH)₂:
Molarity = number of moles / volume in L
Molarity = 0.007 mol /0.015 L
Molarity = 0.47 M
The H₂O is the formula of the chemical specie while the (s), (l) and (g) tell us the state in which it is present, where s is solid, l is liquid and g is gas.
Answer:
Oxygen, bromine, iron, helium
Explanation:
is defined as the standard enthalpy of formation. By definition, the standard enthalpy of formation is equal to 0 kJ/mol for the substances in their standard states, that is, at room temperature and 1 atm pressure.
Simply speaking, looking at the substances given, we need to understand whether their states agree with what we expect to see at standard conditions (e. g., sodium is a metal, fluorine is a gas, bromine is a liquid at standard conditions). Those are substances consisting of just one type of atoms.
- Firstly, oxygen is a gas at standard conditions and it is diatomic, so its
.
- Although nitrogen is a gas at standard conditions, it is diatomic, so
.
- Water is a liquid at standard conditions, however, it consists of two types of atoms, hydrogen and oxygen, so
. - Bromine is a liquid at standard conditions, so .
- Iron is a solid at standard conditions, it's a metal, so .
- Helium is a gas at standard conditions, it belongs to noble gases, so .
- Sulfur is a solid at room conditions, however, the conformation it has is
and not 
, so .
Filtration is a method for separating an insoluble solid from a liquid.
Answer:
The answer is "−847 J/K".
Explanation:
The given expression is:
2Al(s)+ Fe2O3(s) → Al2O3(s)+ 2Fe(s)
Δ
∑(Δ
)
by the above definition Δ
For Such a Component under standard conditions from its standard state, that also applies here. But, we start taking the overview and follow the conventions of signing:
Δ -847 
