H2(g) + I2(g) → 2HI(g) Which of the following will decrease the rate of the forward reaction? A. add some H2 B. remove some HI C. increase the volume D. increase the temperature

h2 {g} would go into the fug so then u tage the oother rate then put the temp so it would be d

Explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the answer for sodium is 0.186

the answer for rubidium is 0.251