Ca3(PO4)2 is the correct formula.
A total of 1.505×10^23 lead atoms
In the lungs, the volume of lead equals the total lung volume, which is 5.60L
1 mole corresponds to 22.4L
Thus, 5.6L of lead converts to 5.6/22.4 = 0.25 moles
According to Avogadro's law
1 mole of lead contains 6.02×10^23 lead atoms
Thus, 0.25 moles of lead equates to 0.25×6.02×10^23 = 1.505×10^23 lead atoms
The correct equation is (C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l). A buffer system is composed of a weak acid and its corresponding salt, effectively stabilizing the pH levels within a solution. The buffer works by adjusting the concentrations of the conjugate acid and base, maintaining the pH constant.
The concentration of the HCl solution can be determined as follows:
The reaction equation is written as
NaOH + HCl = NaCl + H2O
Next, the moles of NaOH are calculated: moles = molarity x volume /1000
= 5 x 2/1000 = 0.01 moles
Using the mole ratio of NaOH to HCl, which is 1:1, the moles of HCl is also equal to 0.01 moles
The concentration is given by: concentration = moles/volume x 1000
= 0.01/10 x 1000 = 1M
Context:
175 kilograms of methane (CH4) is to be converted into hydrogen cyanide (HCN)
The equation that balances this reaction is listed here:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To find the quantities of ammonia and oxygen needed, we will use 175 kg of CH4 as our reference.
Molar masses are as follows:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
For ammonia: mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
This results in 185.94 kg of NH3 required
For oxygen: mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
So the mass of O2 needed equals 525 kg
To derive the mass of oxygen: mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
This gives a mass of O equal to 131.25 kg O
