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3 years ago

Question 1: A substance that is soluble in two liquids and makes an emulsion last longer is called what?

Chemistry

1 answer:

Pavlova-9 [17]3 years ago

3 0

<u>Answer:</u>

1. A substance that is soluble in two liquids and makes an emulsion last longer is called "Emulsifier".

2. The process that reduces the size of particles so emulsions will last longer is called "Homogenization".

<u>Explanation:</u>

Emulsifiers are additives which enable two liquids to mix around each other. Water and oil separate in a container, for an instance, but using an emulsifier can make the liquids blend along. It is widely used on various foods and beverages. Egg yolks and mustard are a few examples of emulsifiers.

Homogenization is the physical mechanism by which the fat molecules in milk are broken down because then they stay incorporated instead of segregated as cream. Majority of the milk sold in the United States is homogenized.

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Draw the structures of all monobromo derivatives of pentane, c5h11br, which contain a 4-carbon chain.

SVETLANKA909090 [29]

Two or more compounds that have same molecular formula but differ in the arrangement of atoms in molecule and thus posses different properties are known as isomers.

The molecular formula of pentane is $C_5H_1_2$ substituting one hydrogen from pentane with bromine results in the formation of monobromo derivatives of pentane having molecular formula, $C_5H_1_1Br$ .

The structure of monobromo derivatives of pentane that is 1-bromopentane, 2-bromopentane, and 3-bromopentane and having molecular formula, $C_5H_1_1Br$ is shown in the image.

The other two arrangements of monobromo derivatives of pentane that is 2-bromo-2-methylbutane and 2-bromo-3-methylbutane is shown in the image.

There are different structures of monobromo derivatives of pentane having molecular formula, $C_5H_1_1Br$ which contain a 4-carbon chain are 1- bromo-2-methylbutane and 1-bromo-3-methylbutane shown in the image.

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3 years ago

What are the two sources of energy for the Earth system?

lozanna [386]

The 2 sources are solar and nuclear energy.

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3 years ago

What does atomic mass on the periodic table represent?

neonofarm [45]

The atomic mass on the periodic table represents the sum of number of protons and number of neutrons.

Atomic mass = Number of protons + number of neutrons

Hope this helps!

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3 years ago

Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1

Drupady [299]

Answer : The value of second ionization energy of Ca is 1010 kJ.

Explanation :

The formation of calcium oxide is,

$Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)$

$\Delta H_f^o$ = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of $CaO$ :

(1) Conversion of solid calcium into gaseous calcium atoms.

$Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)$

$\Delta H_s$ = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

$Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)$

$\Delta H_I_1$ = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

$Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)$

$\Delta H_I_2$ = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

$O_2(g)\overset{\Delta H_D}\rightarrow OI(g)$

$\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)$

$\Delta H_D$ = dissociation energy of oxygen = $\frac{498}{2}=249kJ$

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

$O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)$

$\Delta H_E_1$ = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

$O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)$

$\Delta H_E_2$ = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

$Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)$

$\Delta H_L$ = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L$

Now put all the given values in this equation, we get:

$-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)$

$\Delta H_I_2=1010kJ$

Therefore, the value of second ionization energy of Ca is 1010 kJ.

6 0

3 years ago

A 2.00L flask was filled with 4.00 mol of HI at a certain temperature and given sufficient time to react. At equilibrium the con

inn [45]

Answer:

The equilibrium concentration of I₂ is 0.400 M and HI is 1.20 M, the Keq will be 0.112.

Explanation:

Based on the given information, the equilibrium reaction will be,

2HI (g) ⇔ H₂ (g) + I₂ (g)

It is given that 4.00 mol of HI was filled in a flask of 2.00 L, thus, the concentration of HI will be,

= 4.00 mol/2.00 L

= 2.00 mol/L

Based on the reaction, the initial concentration of 2HI is 2.00, H₂ is 0 and I₂ is O. The change in the concentration of 2HI is -x, H₂ is x and I₂ is x. The equilibrium concentration of 2HI will be 0.200-x, H₂ is x and I₂ is x.

It is given that at equilibrium, the concentration of H₂ or x is 0.400 M.

Now the equilibrium concentration of HI will be,

= 2.00 -2x

= 2.00 - 2 × 0.400

= 1.20 M

The equilibrium concentration of I₂ will be,

I₂ = x

= 0.400 M

The equilibrium constant (Keq) will be,

Keq = [H₂] [I₂] / [HI]²

= (0.400) (0.400) / (1.20)²

= 0.112

Thus, the Keq of the reaction will be 0.112.

5 0

3 years ago