pH=4.625
The classification of this sample of saliva : acid
<h3>Further explanation</h3>
The water equilibrium constant (Kw) is the product of concentration
the ions:
Kw = [H₃O⁺] [OH⁻]
Kw value at 25° C = 10⁻¹⁴
It is known [OH-] = 4.22 x 10⁻¹⁰ M
then the concentration of H₃O⁺:
![\tt 10^{-14}=4.22\times 10^{-10}\times [H_3O^+]\\\\(H_3O^+]=\dfrac{10^{-14}}{4.22\times 10^{-10}}=2.37\times 10^{-5}](https://tex.z-dn.net/?f=%5Ctt%2010%5E%7B-14%7D%3D4.22%5Ctimes%2010%5E%7B-10%7D%5Ctimes%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%28H_3O%5E%2B%5D%3D%5Cdfrac%7B10%5E%7B-14%7D%7D%7B4.22%5Ctimes%2010%5E%7B-10%7D%7D%3D2.37%5Ctimes%2010%5E%7B-5%7D)
pH=-log[H₃O⁺]
Saliva⇒acid(pH<7)
Answer:
0.158 moles
Explanation:
We are given;
9.50 x 10^22 molecules of CO
We are required to determine the number of moles;
We need to know;
1 mole of a compound = 6.022 × 10^23 molecules
Therefore;
9.50 x 10^22 molecules of CO will be equivalent to;
= 9.50 x 10^22 molecules ÷ 6.022 × 10^23 molecules/mole
= 0.158 moles
Therefore, the number of moles are 0.158 moles
Answer:
Explanation:
From the information given:
Mass of carbon tetrachloride = 5 kg
Pressure = 1 bar
The given density for carbon tetrachloride = 1590 kg/m³
The specific heat of carbon tetrachloride = 0.84 kJ/kg K
From the composition, the initial volume of carbon tetrachloride will be:
= 0.0031 m³
Suppose
is independent of temperature while pressure is constant;
Then:
The change in volume can be expressed as:





However; the workdone = -PdV

W = - 7.6 J
The heat energy Q = Δ h


Q = 84 kJ
The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;
ΔU = ΔQ + W
ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)
ΔU = 83.992 kJ
Answer:
The empirical formula is ZnO2
Explanation:
What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?
Step 1: Data given
Suppose the compound has a mass of 100.0 grams
A compound contains:
67.1 % Zinc = 67.1 grams
100 - 67.1 = 32.9 % oxygen = 32.9 grams
Molar mass of Zinc = 65.38 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles of Zinc
Suppose the compound is 100 grams
Moles Zn = 67. 10 grams / 65.38 g/mol
Moles Zn = 1.026 moles
Step 3: Calculate moles of O
Moles O = 32.90 grams / 16.00 g/mol
Moles O = 2.056 moles
Step 4: Calculate mol ratio
We divide by the smallest amount of moles
Zn: 1.026/1.026 = 1
O: 2.056/1.026 = 2
The empirical formula is ZnO2
To control this we can calculate the % Zinc for 1 mol
65.38 / (65.38+2*16) = 0.67.1 = 67.2 %