Answer: 4 molL-1
Explanation:
Detailed solution is shown in the image attached. The number of moles of NaCl is first obtained. Since the molarity must be in units of molL-1, the volume is divided by 1000 and the formula stated in the solution is applied and the answer is given to one significant figure.
When the reaction equation is:
CaSO3(s) → CaO(s) + SO2(g)
we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.
to get the moles of SO2 we are going to use the ideal gas equation:
PV = nRT
when P is the pressure = 1.1 atm
and V is the volume = 14.5 L
n is the moles' number (which we need to calculate)
R ideal gas constant = 0.0821
and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K
so, by substitution:
1.1 * 14.5 L = n * 0.0821 * 285.5
∴ n = 1.1 * 14.5 / (0.0821*285.5)
= 0.68 moles SO2
∴ moles CaSO3 = 0.68 moles
so we can easily get the mass of CaSO3:
when mass = moles * molar mass
and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol
∴ mass = 0.68 moles* 120 g/mol = 81.6 g
Answer:
C₂H₄O₂ and NaC₂H₃O₂ are reactants.
Explanation:
Word equation:
Acetic acid + sodium acetate → sodium diacetate
Chemical equation:
C₂H₄O₂ + NaC₂H₃O₂ → C₄H₇NaO₄
This is a synthesis reaction in which simple reactants combine to form complex product.
This is also balanced chemical equation because there are equal number of atoms of all elements on both side of equation. Thus it follow the law of conservation of mass.
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Answer:a weak acid and a weak base
Explanation:
Answer:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.