Answer is: mass of calcium hydroxide is 46.98 grams.
Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.
m(CaO) = 35.55 g.
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 35.55 g ÷ 56 g/mol.
n(CaO) = 0.634 mol; limiting reactant.
m(H₂O) = 125 mL · 1.000 g/mL.
m(H₂O) = 125 g.
n(H₂O) = 125 g ÷ 18 g/mol.
n(H₂O) = 6.94 mol.
From chemical reaction: n(CaO) : n(Ca(OH)₂) = 1 : 1.
n(Ca(OH)₂) = 0.634 mol.
m(Ca(OH)₂) = 0.634 mol · 74.1 g/mol = 46.98 g.
Because they are Isotopes thats why because isotopes have the same atomic number but diffrent mass number
Should be oxygen .. but water comes out of the leafs stomata ..
<u>Answer:</u> The final temperature of the mixture is 51.49°C
<u>Explanation:</u>
When two samples of water are mixed, the heat released by the water at high temperature will be equal to the amount of heat absorbed by water at low temperature
The equation used to calculate heat released or absorbed follows:
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of water at high temperature = 140 g (Density of water = 1.00 g/mL)
= mass of water at low temperature = 230 g
= final temperature = ?°C
= initial temperature of water at high temperature = 95.00°C
= initial temperature of water at low temperature = 25.00°C
c = specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
![140\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]](https://tex.z-dn.net/?f=140%5Ctimes%204.186%5Ctimes%20%28T_%7Bfinal%7D-95%29%3D-%5B230%5Ctimes%204.186%5Ctimes%20%28T_%7Bfinal%7D-25%29%5D)
Hence, the final temperature of the mixture is 51.49°C
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
