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Please answer in the picture chemistry

look at the graph

Explanation:

We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.

Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.

So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.

Answer:             1.0  × 10⁻⁵ msolution:                     as we know ph is given as,                                    ph  = -log [h⁺]and,                                    [h⁺]  =  10^-ph    (1)putting value of ph in eq. 1,                                    [h⁺]  =  10⁻⁵                                    [h⁺]  =  1.0  × 10⁻⁵ m