Answer:
if the object sank then that object has a greater density then water. if the object floated then its density is lower then water.
Explanation:
lets say object 1 has a density of 24/cm3. the density is greater then water (1.0000g/cm3) so it would sink. now lets say object 2 has a density of 0.79383g/cm3 since it's less then the density of water (1.0000g/cm3) it would float.
A chemical reaction of exothermic kind releases energy in the form of heat and light.
We are given that reactant releases 27.4kJ/g for every gram of reactant consumed.
We are required to find number of grams of reactant that has been consumed
It is also given that 880J of energy is released
Hence, grams of reactants consumed is 880/27.4gm
Disclaimer:
It has been asked by the student to leave the question as mathematical expression.
For further reference:
brainly.com/question/15000187?referrer=searchResults
#SPJ9
Answer:
ΔHrxn = 239 kj/mol
Explanation:
Given data:
Mass of sodium = 0.230 g
Heat produced = 2390 J
Solution:
Chemical equation:
2Na + 2HCl → 2NaCl + H₂
Number of moles of sodium:
Number of moles of sodium = mass / molar mass
Number of moles of sodium = 0.230 g / 23 g/mol
Number of moles of sodium = 0.01 mol
Enthalpy of reaction:
ΔHrxn = 2390 J / 0.01 mol
ΔHrxn = 239000 j/mol
ΔHrxn = 239 kj/mol
Answer:
The final molarity of acetate anion in the solution is 0.0046 moles
Explanation:
The balanced equation is
Cu(C₂H₃O₂)₂ + Na₂CrO₄ = CuCrO₄ + 2Na(C₂H₃O₂)
Therefore one mole of Cu(C₂H₃O₂)₂ react with one mole of Na₂CrO₄ to form one mole of CuCrO₄ and two moles of Na(C₂H₃O₂)
Mass of copper (II) acetate present = 0.708 g
Volume of aqueous sodium present = 50 mL
Molarity of sodium chromate = 46.0 mM
Therefore
Number of moles of sodium chromate present = (50 mL/1000)×46/1000 = 0.0023 M
Number of moles of copper (II) acetate present = 181.63 g/mol
number of moles of copper (II) acetate present = (0.708 g/181.63 g/mol) =0.0039 moles
Therefore 0.0039 moles of Cu(C₂H₃O₂)₂ × (2 moles of Na(C₂H₃O₂))/1 Cu(C₂H₃O₂)₂) = 0.00779 moles of Na(C₂H₃O₂)
also 0.0023 moles of Na₂CrO₄ × (2 moles of Na(C₂H₃O₂))/1 Na₂CrO₄) = 0.0046 moles of Na(C₂H₃O₂)
Therefore the Na₂CrO₄ is the limiting reactant and 0.0046 moles of Na(C₂H₃O₂) or acetate anion is formed
