The equation we use is F=ma
(Force = mass x acceleration)
We are going to put each number in the equation:
F=ma
20N = 5kg x (?)
So, to find a (acceleration), we do F/m
[acceleration = force/mass]
a = 20/5 = 4
The answer is 4m/s^2
Answer:
Q₁- The concentration of HCl = 0.075 N = 0.075 M.
Q₂- The concentration of KOH = 7.675 mN = 7.675 mM.
Q₃- The concentration of H₂SO₄ = 0.2115 N = 0.105 M.
Q₄- The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
Explanation:
<u><em>Q₁:
</em></u>
- As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
- The normality of the NaOH and HCl = Their molarity.
∵ (NV)NaOH = (NV)HCl
∴ N of HCl = (NV)NaOH / (V)HCl = (0.15 N)(67 mL) / (134 mL) = 0.075 N.
∴ The concentration of HCl = 0.075 N = 0.075 M.
<em><u>Q₂:</u></em>
- As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
- The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.
∵ (NV)H₂SO₄ = (NV)KOH
∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.
∴ The concentration of KOH = 7.675 mN = 7.675 mM.
<em><u>Q₃:</u></em>
- As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
- The normality of NaOH = Molarity of NaOH = 0.5 N.
∵ (NV)H₂SO₄ = (NV)NaOH
∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄ = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.
∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.
<em><u>Q₄:</u></em>
- The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
- The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
- End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
- The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
Mole percent of O2 = 10% = 0.1 moles
Mole percent of N2 = 10% = 0.1 moles
Mole percent of He = 80% = 0.8 moles
Molar Mass of O2 = (2 x 16) x 0.1 = 3.2
Molar Mass of N2 = (2 x 14) x 0.1 = 2.8
Molar Mass of He = 4 x 0.8 = 3.2
1. Molar Mass of the mixture = 3.2 + 2.8 + 3.2 = 9.2 grams
2. Since at constant volume density is proportional to mass, so the ratio of
mass will be the ratio of density.
Ratio = Molar Mass of the mixture / Molar Mass of O2 = 9.2 / 32 = 0.2875
The reaction used to produce sugar in plants is Photosynthesis,
6CO₂ + 6H₂O + heat ⇆ C₆H₁₂O₆ + 6O₂↑
we can see that this reaction is endothermic, it absorbs heat to occur
If we were to add more heat or increase the temperature, the left side of the equilibrium will be able to react much more, which would produce more glucose (aka sugar)
Hence increasing the temperature will shift the equilibrium to the right or towards the Products
