What is Entropy?

A. The opposite of Enthalpy
B. The temperature at which a reaction is feasible.
C. A measure of disorder.
D. Free energy used for a reaction.

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Answer:

1.) Propanone (ketone)

2.) Ethanal( aldehyde)

3.) 3-phenyl-2-propenal (aldehyde)

4.) Butanone (ketone)

5.) Ethanol ( alcohol)

6.) 2-propanol (alcohol)

Explanation:

In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.

Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.

Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal

Answer:

Molar mass for eugenol is 161.3 g/mol

Explanation:

This question talks about freezing point depression:

Our solute is eugenol.

Our solvent is camphor.

Formula to state the freezing point depression difference is:

ΔT = Kf . m . i where

ΔT = Freezing T° of pure solvent - Freezing T° of solution

In this case ΔT = 0.62°C

Kf for camphor is: 37°C /m

As eugenol is an organic compund, i = 1. No ions are formed.

To state the molar mass, we need m (molal)

Molal are the moles of solute in 1kg of solvent. Let's replace data:

0.62°C = 40 °C/m . m . 1

0.62°C / 40 m/°C = 0.0155 m

We convert mass of camphor from g to kg = 100 g . 1kg / 1000g = 0.1 kg

0.0155 molal = moles of solute / 0.1 kg

0.0155 m/kg . 0.1 kg = 0.00155 moles

We know that these moles are contained in 250 mg, so the molar mass will be:

0.25 g / 0.00155 mol = 161.3 g/mol

Notice, we convert mg to g, for the units!

Answer:

KOH + HClO4 = K(ClO4) + H2O

Explanation:

KClO4 + H2O : mix and match the letters if you need to guess

Explanation:

Iron has 2 atoms and 3atoms.

So,X=2,3

Answer:

209

83 Bi

Explanation:

213 213 - 4 4

85 At = 85 - 2 Y + 2 He

Given question is:

Given a food with a relative humidity of 81%, what is the water activity in that food?

Explanation:

Water activity in food can be determined by using the below-shown formula:

[tex]water activity=\frac{equilibrium relative humidity}{100}[/tex]

Equilibrium is established between the vapor pressure of food and the surrounding air media.

Thus, for the given food the relative humidity is 81%.

hence, its water activity is

[tex]81/100\\=0.81[/tex]

Answer:

gravity, ground, friction, rolling resistance, and air resistance.

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

Answer:

% yield of reaction is 26.4

Explanation:

The reaction is:

Mg + I₂ →  MgI₂

Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:

10 g . 1mol / 24.3 g = 0.411 moles of Mg

95 g . 1mol / 253.8g = 0.374 moles of I₂

Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine

For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.

As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂

We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g

To calculate the percent yield:

% yield = (yield produced /theoretical yield) . 100

% yield = (27.5 g/ 104g) . 100 = 26.4

Answer:

4.4 mL

Explanation:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed?

Step 1: Write the balanced equation

CH₄(g) + 4 Cl₂(g) ⇒ 4 HCl(g) + CCl₄(g)

Step 2: Establish the appropriate volume ratio

For gases under the same conditions, the volume ratio is equal to the molar ratio. The molar ratio of CH₄ to HCl is 1:4.

Step 3: Calculate the volume of HCl produced from 1.1 mL of CH₄

1.1 mL CH₄ × 4 mL HCl/1 mL CH₄ = 4.4 mL HCl

[tex] \boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }[/tex]

[tex]\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}[/tex]

= 2 × 23 + 2 × 52 + 2 × 16

= 182 grams

1 mole of [tex]Na_2Cr_2O_2[/tex] weighs = 182 g

8 moles weigh = 8× 182

=[tex] \mathfrak{\blue {\boxed{\underline {1456 \: grams}}}} [/tex]

or

[tex] \mathfrak{\blue {\boxed{\underline {1. 46 \:kg }}}} [/tex]

Answer:

[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

Answer:

hydrogen bonding

Explanation:

just took the test :D

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

∴

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

QUESTION:- Which particle has a mass of 9.11 x 10^-28g and charge of -1?

OPTIONS:-

A. electron

B. proton

C. neutron

ANSWER:-

CHARGE ON PROTRON IS +1 AND IT HAS MASS OF [tex]1.6 \times 10 {}^{ - 27} [/tex] SO IT CANNOT BE URE ANSWER

THERE IS NO CHARGE ON NEUTRON AND HAS MASS ALMOST EQUAL TO THE PROTON SO IT ALSO CANNOT BE URE ANSWER

MASS OF THE ELECTRON:- [tex]9.11 \times 10^{ - 28} [/tex]

CHARGE ON ELECTRON:- [tex] -1[/tex]

SO URE ANSWER IS ELECTRON

Answer:

[tex]\alpha=17.7[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=13g[/tex]

Volume [tex]V=10mL[/tex]

Angle [tex]\theta=23[/tex]

Sample Tube=10cm

Generally the equation for concentration is mathematically given by

 [tex]C=m/v[/tex]

 [tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]

Therefore the Specific Rotation

 [tex]\alpha=frac{\theta }{m*l}[/tex]

 [tex]\alpha=frac{23 }{1.3*1.0}[/tex]

 [tex]\alpha=17.7[/tex]

Answer:

lead

Explanation:

All metals are good conductors of heat and electricity. This property allows the heat to flow through them. The metals which are used to make cookware are Aluminum, Copper and Iron. The correct options are A, B and C.

The metals are defined as the substances which are formed naturally below the surface of the earth. Most of the metals are lustrous or shiny. They are very strong and durable, so they are used to make many things like satellites, automobiles, cooking utensils, etc.

Malleability is the property of metals which allows them to be beaten into flat sheets. Aluminium sheets are used in the manufacture of aircrafts and utensils because of its light weight and strength.

So here aluminium, copper and iron are used to make utensils, since they are good conductors of heat and electricity.

Thus the correct options are A, B and C.

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Answer:

57------6.9

330------.87

Explanation:

Answer:

The required volume of air is 3.64 L

Explanation:

Ideal gas equation:-

The relation between pressure, volume, and temperature of the gas is known as the ideal gas equation and it is given as,

   PV=RT

Where R=gas constant

Now,

Let the volume of air is, V

According to the question we have

Temperature, T= 60°C= (60+273) K= 333 K

Atmospheric pressure, P= 760 mm

And gas constant, R= 8.314 J/mole K

Substitute values in ideal gas equation and we get

  760V= 8.314(333)

  V= 2768.562/760

  V= 3.642 L

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Answer:

The correct answer is - 2.557 KJ

Explanation:

In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.

we can calculate this energy by the following formula:

Q = met

where, m = mass,

e = specific heat

t = temperature

then,

Q = 78*0.14* (273-38.8)

here 0.14 = C(Hg)

= 2.557 Kj

Answer:

C. By super-cooling certain types of plasma.

Explanation:

Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.

Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.

When certain types of plasma are super cooled, Bose-Einstein condensate are formed.

Answer:

NH3

Explanation:

It’s a polyatomic ion. You’ll just have to memorize it!

Answer:

NH3

Explanation:

Ammonia is a compound of nitrogen and hydrogen with the formula NH3

Answer:

addition of HNO3 HNOX3 to Cu - Aueous

addition of H2SO4 HX2SOX4 to CuO - Aqueous

addition of Z n Zn to C u S O 4 CuSOX4 - Solid

addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid

heating of C u ( O H ) - Solid

Explanation:

Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.

Answer:

The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"

Explanation:

Calculating the rate of the equation:

[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]

Rate:

[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]

Answer:

26.6

Explanation:

Step 1: Calculate the molar concentrations

We will use the following expression.

M = mass solute / molar mass solute × liters of solution

[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M

[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M

[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M

Step 2: Make an ICE chart

        CO(g) + 2 H₂(g) ⇄ CH₃OH(g)

I        0.184      0.227           0

C         -x           -2x             +x

E     0.184-x   0.227-2x        x

Since [CH₃OH]e = x, x = 0.0523

Step 3: Calculate all the concentrations at equilibrium

[CO]e = 0.184-x = 0.132 M

[H₂]e = 0.227-2x = 0.122 M

[CH₃OH]e = 0.0523 M

Step 4: Calculate the equilibrium constant (Kc)

Kc = [CH₃OH] / [CO] [H₂]²

Kc = 0.0523 / 0.132 × 0.122² = 26.6

Answer:

0.169 mole of Al

Explanation:

We'll begin by by calculating the number of mole in 45 gof AlBr₃. This can be obtained as follow:

Mass of AlBr₃ = 45 g

Molar mass of AlBr₃ = 27 + (3×80)

= 27 + 240

= 267 g/mol

Mole of AlBr₃ =?

Mole = mass /molar mass

Mole of AlBr₃ = 45 / 267

Mole of AlBr₃ = 0.169 mole

Finally, we shall determine the number of mole of Al needed to produce 45 g (i.e 0.169 mole) of AlBr₃. This can be obtained as illustrated below:

2Al + 3Br₂ —> 2AlBr₃

From the balanced equation above,

2 moles of Al reacted to produce 2 moles of AlBr₃.

Therefore, 0.169 mole of Al will also react to produce 0.169 mole of AlBr₃.

Thus, 0.169 mole of Al is needed for the reaction.

Answer:

this product can be achieved using the starting material shown is by use of NaOH as catalyst.

Answer:

By using NaOH as catalyst.

Explanation:

This product can be achieved using the starting material shown is by the use of the NaOH as catalyst.

Answer:

There are 8 protons

Explanation:

[tex]{ \bf{atomic \: number = proton \: number}}[/tex]