The fusion on water is an enxothermic change but at this temp, the temp-entropy product is outweighed by the change in enthalpy.
D i took the test Enjoy :))))
Answer:
108.6 g
Explanation:
- 2NaN₃(s) → 2Na(s) + 3N₂(g)
First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 0 °C ⇒ 0 + 273.2 = 273.2 K
<u>Inputting the data</u>:
- 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 2.5 mol N₂ *
= 1.67 mol NaN₃
Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:
- 1.67 mol * 65 g/mol = 108.6 g
12 moles, there is 3 moles of copper in one mole of Cu3 (PO4)2, there four multiple 3 by 4 and you get 12
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
