Answer:
Explanation:
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In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:
For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:
Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:
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Answer:
11·699
Explanation:
Given the concentration of hydroxide ion in the solution is 5 × 
M
Assuming the temperature at which it is asked to find the pH of the solution be 298 K
<h3>At 298 K the dissociation constant of water is
</h3><h3>∴ pH + pOH = 14 at 298 K</h3><h3>pOH of the solution = -log( concentration of hydroxide ion )</h3>
∴ pOH of the given solution = - log(5 × = -0·699 + 3 = 2·301
pH of the given solution = 14 - 2·301 = 11·699
∴ pH of the solution = 11·699
Answer:
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Answer:
As you go down a group on the periodic table, you get more electron shells and therefore a larger atomic radius.
Answer:
V = 48.64 L
Explanation:
Given data:
Mass of CO₂ = 85.63 g
Temperature = 273 K
Pressure = 1 atm
Volume of CO₂ = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 85.63 g / 40 g/mol
Number of moles = 2.14 mol
Volume in Litter:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
1 atm× V = 2.14 mol ×0.0821 atm.L/mol.K ×273 K
V = 48.64 atm.L / 1 atm
V = 48.64 L
