Molarity = moles of solute/volume of solution in liters.
The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.
(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.
Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:
(0.8017 moles NaCl)/(2.2 L) = 0.364 M.
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
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Answer:
1.3×10⁻³ M
Explanation:
Hello,
In this case, given the dissociation reaction of acetic acid:
We can write the law of mass action for it:
![Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BCH_3CO_2%5E-%5D%7D%7B%5BCH_3CO_2H%5D%7D)
Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change 
due to the dissociation extent, we are able to rewrite it as shown below:
Thus, via the quadratic equation or solve, we obtain the following solutions:
Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.
Regards.
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The pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
<h3>What is pH? </h3>
pH is defined as the concentration of the hydrogen bond which is released or gained by the species in the solution which depicts the acidity and basicity of the solution.
<h3>What is pOH? </h3>
pOH is defined as the concentration of the hydronium ion present in solution.
pOH value is inversely proportional to the value of pH.
pH value increases, pOH value decreases and vice versa.
Given,
Total H+ ions = 2.95 ×10^(-12)M
<h3>Calculation of pH</h3>
pH = -log[H+]
By substituting the value of H+ ion in given equation
= log(2.95× 10^(-12) )
= 13.5
Thus we find that the pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
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