Answer: Resulting solution will not be neutral because the moles of $OH^-$ ions is greater. The remaining concentration of $[OH^-]$ ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

$[HNO_3]$ = 0.200 M

$[Ca(OH)_2]$ =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of $H^+$ ion from both the acids

moles of $H^+$ in HCl

$HCl\rightarrow {H^+}+Cl^-$

if 1 L of $HCl$ solution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 $\times$ 0.1 moles= 0.005 moles (1L=1000mL)

moles of $H^+$ in HCl = 0.005 moles

Similarliy

moles of $H^+$ in $HNO_3$

$HNO_3\rightarrow H^++NO_3^-}$

If 1L of $HNO_3$ solution= 0.200 moles

Then 0.1L of $HNO_3$ solution= 0.1 $\times$ 0.200 moles= 0.02 moles

moles of $H^+$ in $HNO_3$ =0.02 moles

so, Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles .....(1)

Step 2: Calculating the total moles of $[OH^-]$ ion from both the bases

Moles of $OH^-\text{ in }Ca(OH)_2$

$Ca(OH)_2\rightarrow Ca^2{+}+2OH^-$

1 L of $Ca(OH)_2$ = 0.0100 moles

Then in 0.5 L $Ca(OH)_2$ solution = 0.5 $\times$ 0.0100 moles = 0.005 moles

$Ca(OH)_2$ produces two moles of $OH^-$ ions

moles of $OH^-$ = 0.005 $\times$ 2= 0.01 moles

Moles of $OH^-$ in $RbOH$

$RbOH\rightarrow Rb^++OH^-$

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 $\times$ 0.100 moles = 0.02 moles

Moles of $OH^-$ = 0.02 moles

so,Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles ....(2)

Step 3: Comparing the moles of both $H^+\text{ and }OH^-$ ions

One mole of $H^+$ ions will combine with one mole of $OH^-$ ions, so

Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles....(1)

Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of $H^+$ ions = total moles of $OH^-$ ions

0.025 moles $H^+$ will neutralize the 0.025 moles of $OH^-$

Moles of $OH^-$ ions is in excess (from 1 and 2)

The remaining moles of $OH^-$ will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of $OH^-$ ions = $\frac{\text{Moles remaining}}{\text{Total volume}}$

$[OH^-]=\frac{0.005}{0.85}=0.0058M$

6 0

3 years ago

A mixture with medium sized particles that do not settle out is called

Alexandra [31]

Answer: colloid

Explanation:

Colloids are solutions in which small sized particles are suspended throughout the solution as they do not settle own their own. Colloids are defined as the mixtures where the size of the particle is within the range of 2nm to 1000 nm. In these mixtures, physical boundary is seen between the dispersed phase and dispersed medium.

Colloids are solutions with particle size intermediate between true solutions and suspensions. Suspensions have large sized particles which settle when left undisturbed for sometime and thus can be filtered off easily. The particle size in colloids is less and hence they do not settle under the effect of gravity.

7 0

3 years ago

The number of molecules in 7g nitrogen​

The number of molecules in 7g nitrogen