Answer:
C. Cornea and Lens
Explanation:
Cornea This is the front part of the tough outer coat that is transparent. It refracts light (bends it as it enters the eyes to ensure it is in the right place).
Lens A transparent, biconvex, flexible disc behind the iris. Focuses the light onto the retina ( the light sensitive part of the eye)
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.
The answer is 35,000 calories
8S or S8 but it is not 8S8
8S8 does not make senes
Beside, is this element is existed.
It means that u have 64 molecules of Sulphur.
Answer : The final equilibrium temperature of the water and iron is, 537.12 K
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of iron = 560 J/(kg.K)
= specific heat of water = 4186 J/(kg.K)
= mass of iron = 825 g
= mass of water = 40 g
= final temperature of water and iron = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:
Therefore, the final equilibrium temperature of the water and iron is, 537.12 K
