The solution of weak monoprotic acid which is 0.10 M has pH = 3 .
<h3>Are strong and weak acids both monoprotic?</h3>
The greater the Ka value, the stronger the acid. Strong acids dissociate totally into protons and their respective conjugate bases. Diprotic acids have two acid dissociation steady values Ka.
<h3>How do you identify a monoprotic acid?</h3>
A monoprotic acid can be recognized via the presence of an equivalence point in a titration curve. Titration is a method used to decide the unknown attention of a solution from a answer whose awareness is known.
Learn more about monoprotic acids here:
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Answer:
C. Bent
Explanation:
Only a bent molecule is polar.
The two bond dipoles do not cancel, so the molecule is polar.
A is wrong. In a linear molecule, the bond dipoles are equal and in opposite directions, so they cancel. The bonds are polar, but the molecule is nonpolar.
B is wrong. In a trigonal planar molecule like that shown below, the resultant of the bond dipoles of the two lower bonds is equal to the upper bond dipole and in the opposite direction. The two dipoles cancel, so the molecule is nonpolar.
D is wrong. A tetrahedral molecule may have polar bonds, but it is a nonpolar molecule. The two bonds on the right have a net dipole to the left, and the two bonds on the left have an equal resultant to the right. The dipoles cancel, so the molecule is nonpolar.
Answer:
mix 1 mL of the stock solution of MgCl2, 300 mL of the stock solution of NaCl and 699 mL of water.
Explanation:
We need to determine the volume necessary of both stock solutions. When a dilution is made, a certain volume of the stock solution is collected and then more solvent is added to it until the volume is completed. So, the number of moles of the solute in both solutions are the same, and it is the concentration (C) multiplied by the volume (V). If 1 is the stock solution and 2 the diluted solution:
C1*V1 = C2*V2
So, in this case, the two solutions will be mixed, and then the volume will be completed with the solvent. So, for MgCl2:
C1 = 1.0 M
C2 = 1.0 mM = 0.001 M
V2 = 1.0 L = 1000 mL
1*V1 = 0.001*1000
V1 = 1 mL of the stock solution of MgCl2
For NaCl:
C1 = 0.50 M
C2 = 0.15 M
V2 = 1000 mL
0.50*V1 = 0.15*1000
V1 = 300 mL
So, it will be necessary 1 mL of the stock solution of MgCl2, 300 mL of the stock solution of NaCl and 699 mL of water.
X volume= 2x2x2=8
y volume= 8x8x8=512
512/8=64 so you can fit 64