melting point because the melting point is the heat required to make an object melt anything less than this poitn will not make the substence melt.
Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
Covalent bonds form when electrons are shared between two nonmetals.
Answer:
Their positive charge is located in the small nucleus
Explanation:
Ernest Rutherford performed the gold foil experiment in 1911 where he used alpha particles generated from a radioactive source to bombard a thin gold foil.
In his experiment, he observed that the bulk of the alpha particles passed through the gold foil, just a tiny fraction was deflected back. To explain his findings, Rutherford proposed that an atom is made of positively charged centre where nearly all the mass is concentrated called nucleus. Surrounding the nucleus is a large space containing electrons.
Burette is a very accurate measuring instrument when adding solutions and has a measurement error of 0.05 mL.
Small volumes of solutions can be transferred from the burette at a controllable rate.
In this instance NaOH is in the burette.
Initial reading of NaOH is 0.20 mL
end point is the point at which the chemical reaction reaches completion. In acid base reactions, end point is when all the H⁺ ions have reacted with OH⁻ ions.
final reading of NaOH is 24.10 mL
to find the volume of NaOH dispensed we have to find the difference between final reading and initial reading
volume of NaOH added = 24.10 mL - 0.20 mL = 23.90 mL
volume of NaOH dispensed is 23.90 mL
