Please Help with 2. Thanks

What is potential or potenz hydrogen?

Mrrafil [7]

Answer:

potential or pontenz Hydrogen is the negative logarithm of molar hydrogen ion concentration.

Explanation:

potential Hydrogen or potenz Hydrogen stands for pH

potenz is in german

${ \tt{pH = - log [H {}^{ + } ]}}$

7 0

3 years ago

If 27.50 mL of 0.120 M NaOH neutralizes 0.248 g of HAA, what is the molar mass of unknown amino acid

tino4ka555 [31]

Answer:

75.15 g/mol

Explanation:

First, let us look at the equation of reaction;

$NaOH + HAA --> NaAA + H_2O$

From the balanced equation of reaction, 1 mole of NaOH is required to completely neutralize 1 mole of HAA.

Recall that: mole = molarity x volume.

Therefore, 27.50 mL, 0.120 M NaOH = 0.0275 x 0.120 = 0.0033 moles

0.0033 mole of NaOH will therefore requires 0.0033 moles of HAA for complete neutralization.

In order to find the molar mass of the unknown amino acid, recall that:

<em>mole = mass/molar mass</em>, hence, <em>molar mass = mass/mole</em>.

Therefore, molar mass of HAA = 0.248/0.0033 = 75.15 g/mol

6 0

3 years ago

A thin-walled sphere rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic ener

ICE Princess25 [194]

Explanation:

The kinetic energy of translation

$E_1=\frac{1}{2}mv^2$

m= mass v= linear velocity

The kinetic energy of rotation

$E_2=\frac{1}{2}I\omega^2$

I= MOI of the thin walled sphere =kmR^2

where ω= v/R= angular velocity

$E_2=\frac{1}{2}kmR^2\frac{v}{R}^2$

Then

$\frac{E_1}{E_2} = \frac{0.5mv^2}{0.5kmR^2\frac{v}{R}^2 }$

=1/k

solid sphere: k=0.4; E1/E2 =1/0.4 = 2.5;

hollow sphere: k=2/3; E1/E2 = 1.5

3 0

3 years ago

Kat is investigating a compound and sees that it has even stronger hydrogen bonds than water. What can kat conclude is most like

Sidana [21]

Answer:

It is higher than that of water

Explanation:

Because we now know that through experimentation, the new compound has a higher and stronger hydrogen bonds than water, the specific heat capacity will be higher.

Specific heat capacity is the amount of heat needed to raise the temperature of a unit mass of as substance by 1°C.

This property is a physical property of matter .
Most physical properties are a function of intermolecular forces in a compound.
Since hydrogen bond is a very strong intermolecular force, the specific heat capacity will be stronger for the compound discovered.
This implies that it will require more heat to raise the temperature of a unit mass of this compound by 1°C.

6 0

3 years ago

Read 2 more answers

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(c) $Ksp=2.27x10^{-12}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

(C) $NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

4 years ago