use the factor theorem to determine whether (x-3) is a factor of f(x)=x^4+12x^3+6x+27

use the factor theorem to determine whether (x-3) is a factor of f(x)=x^4+12x^3+6x+27

Answer

(x)=x  ⁴  given that x-3 is a factor of expression

use remainder theorem

put x  - 3  = 0

we get x = 3

plug 3 in place of x in the given value of f (x)

f(x)=x⁴+ 12x³ + 6x + 27

f(3)=(3)⁴+12(3)³+6(3)+27.

     = 81+ 324 + 18 + 27

     = 430

here 430 is not equal to 0 so that  Hence x -3 is not a factor of f(x)=x^4+12x^3+6x+27

A coffee cup calorimeter contains 0.0375 mol HCl dissolved in 1.50 x 10^2 mL of water at 20.51 C. When 1.75 g of solid zinc metal is placed in the calorimeter, the temperature rises to 29.02 C. Assume that no heat is lost to the surroundings.

A coffee cup calorimeter contains 0.0375 mol HCl dissolved in 1.50 x 10^2 mL of water at 20.51 C. When 1.75g of solid zinc metal is placed in the calorimeter, the temperature rises to 29.02 °C.  Assume that no heat is lost to the surroundings.

Write the balanced equation for the reaction that takes place in the calorimeter.  1pt

Balanced reaction

Zn (s)  + 2 HCl (aq) --> ZnCl₂(aq) + H2 (g)

Find the limiting reactant. Show your work. 2 pts.

Number of moles of Zn             = given mass of Zn / molar mass of Zn

                                                  = 1.75 g Zn / 65.3.9 g/mol

                                                  = 0.02676 moles Zn

In equation 1 mole of Zn react with 2 mol of HCl

Hence  0.02676 moles Zn react with        = 2  x 0.02676

                                                             = 0.05352 mol of HCl 

But required moles of HCl (0.05352 mol) > given moles (0.0375)

Hence HCl is limiting reagent

Find the amount of heat lost or gained by the calorimeter. Show your work. 2pts.

Formula of heat change, ∆T       = T₂ - T₁ = 29.02 - 20.51  = 8.51 ° C

Specific heat capacity for water , C = 4.184 J/g-C

Mass of water m                            = 1.50 x 10²  ml = 150 g  (for water 1 ml = 1 g )

Use formula of enthalpy change ∆H = m * C *∆T

Plug the values

                                               ∆H = 150 g  x 4.184 J/g-C  x 8.51 C

∆H = 5341 Joules         

Divide by 1000 to convert in kJ 

                                                ∆H = 5.34 KJ

Temperature is increasing so that it is a exothermic reaction

And for all exothermic reaction ∆H is always negative 
energy lost by reaction and gained by solution.  hence calorimeter gain this energy

 So final answer will ∆H = - 5.34 KJ

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i)              x + y = 5, 2x + 2y = 10

Convert the equation in form of 

a1x + b1y + c1 = 0  and a2x+ b2y + c2 = 0

We get

               x + y  -5 = 0, 2 x + 2 y  -10 = 0

Compare the equation with

              

We get

a1 = 1,            b1       = 1,                 and c1 = -5

a2 = 2             b2        = 2                   and c2 = - 10

                                    

 So we get 

                                  

Dependent and consistent  so it will have many solutions .

(ii)           x – y = 8, 3x – 3y = 16

Convert the equation in form of

a1x + b1y + c1 = 0  and a2x+ b2y + c2 = 0

We get

   x – y -8 = , 3x – 3y -16 = 0

Compare the equation with

                         

We get

a1 = 1,  b1     = - 1,              and c1 = - 8

a2 = 3  b2      =  - 3               and c2 = - 16

 

So we get 

  

So both lines are Inconsistent

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

Compare the equation with

              

We get

a1 = 2,            b1       =  1,                and c1 = - 6

a2 = 4             b2        =  - 2               and c2 = - 4

 

So we get 

So both lines are consistent

Solve the equations graphically

2x + y – 6 = 0,

Subtract 2x and add 6 both side we get

 y  = 6 – 2x

plug x = 1,2 and 3 we get

y         =  6 - 2 * 1      = 4

 y         = 6 -  2 * 2      = 2

y          =  6 -  2 * 3     = 0

X	1	2	3
y	4	2	0

 4 x – 2 y – 4 = 0

Divide by 2 we get

2 x – y - 2                  = 0 

Add y both side we get

2 x – 2            = y

or

y          = 2x – 2

plug x  = 1.2 and 3 we get

y          = 2 * 1 – 2      = 0

y          = 2 * 2 – 2      = 2

y          = 2 * 3 – 2      =  4

X	1	2	3
y	0	2	4

Answer x = 2 and y = 2

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Compare the equation with

         

We get

a1 = 2,            b1       = - 2,              and c1 = - 2

a2 = 2             b2        =  - 2               and c2 = - 5

So both lines are Inconsistent