Ask question

Ask question

All categories

3 years ago

9

Bases dissolve _ and _. 1. sugars 2. fats 3. oils or 4. proteins

1 answer:

geniusboy [140]3 years ago

6 0

Answer:due to principle "like dissolve like" as base is electronegativty it may dissolve with protein at it's isoelectronic point

You might be interested in

21) Scientists represent elements by using<br> a formulas b. subscripts<br> c. symbols<br> d. words

Hunter-Best [27]

They represent elements by using symbols

4 0

2 years ago

A 51.24-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be d

g100num [7]

Answer:

0.40 L

Explanation:

Calculation of the moles of $Ba(OH)_2$ as:-

Mass = 51.24 g

Molar mass of $Ba(OH)_2$ = 171.34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{51.24\ g}{171.34\ g/mol}$

$Moles= 0.2991\ mol$

Volume = 1.20 L

The expression for the molarity is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M$

Thus,

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$Molarity_{working\ solution}=0.100\ M$

$Volume_{working\ solution}=1\ L$

$Volume_{stock\ solution}=?$

$Molarity_{stock\ solution}=0.24925\ M$

So,

$0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}$

$Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L$

<u>The volume of 0.24925M stock solution added = 0.40 L </u>

6 0

2 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

Which type of structure do most ionic compounds form

Setler79 [48]

Answer: crystalline solid

Explanation: Most ionic compounds exist in crystalline solid form especially at room temperature.

4 0

2 years ago

What is the value of x in the equation below?

Colt1911 [192]

If the question includes x only than it’s value would be 1

8 0

3 years ago

Read 2 more answers