This problem is simply converting the concentration from molality to molarity. Molality has units of mol solute/kg solvent, while molarity has units of mol solute/L solution.
2.24 mol H2SO4/kg H2O * (0.25806 kg H2SO4/mol H2SO4) = 0.578 kg H2SO4/kg H2O
That means the solution weighs a total of 1 kg + 0.578 kg = 1.578 kg. Then, convert it to liters using the density data:
1.578 kg * (1000g / 1kg) * (1 mL/1.135 g) = 1390 mL or 1.39 L.
Hence, the molarity is
2.24/1.39 = 1.61 M
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
Answer:
4.75 is the equilibrium constant for the reaction.
Explanation:
Equilibrium concentration of reactants :
![[CO]=0.0590 M,[H_2O]=0.00600 M](https://tex.z-dn.net/?f=%5BCO%5D%3D0.0590%20M%2C%5BH_2O%5D%3D0.00600%20M)
Equilibrium concentration of products:
![[CO_2]=0.0410 M,[H_2]=0.0410 M](https://tex.z-dn.net/?f=%5BCO_2%5D%3D0.0410%20M%2C%5BH_2%5D%3D0.0410%20M)
The expression of an equilibrium constant is given by :
![K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)
4.75 is the equilibrium constant for the reaction.
Answer:
solid
Explanation:
Melting and boiling points of Group 7 elements State at room temperature Room temperature is usually taken as being 25°C. At this temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine and astatine are solids. There is therefore a trend in state from gas to liquid to solid as you go down the group.
Vertical columns are called groups, or families.
They share similar chemical properties.
Answer: <span>have similar chemical properties</span>
