Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
Answer:
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
![K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BAg%24%5E%7B%2B%7D%24%5D%24%5E%7B2%7D%24%5BCO%24_%7B3%7D%5E%7B2-%7D%24%5D%7D%20%3D%20%282x%29%5E%7B2%7D%5Ctimes%200.00750%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5C0.0300x%5E%7B2%7D%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5Cx%5E%7B2%7D%20%3D%202.70%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cx%20%3D%20%5Csqrt%7B2.70%20%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%20%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20maximum%20concentration%20of%20Ag%24%5E%7B%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B-5%7D%7D%5Ctextbf%7B%20mol%2FL%20%7D%7D%24%7D)
Carbonic anhydrase speeds up<span> the transfer of carbon dioxide from cells to the blood.
Hope this work cuz</span>
Answer:
8510 mol
Explanation:you must divide both of the molar masses into each other
1 molecule of glucose contains 6 atoms of C, 12 atoms of H , and 6 atoms of 0.1 mole of glucose contains 6 moles of C atoms , 12 moles of H atoms , and 6 moles of O atoms .
