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3 years ago

F(x) = x3 + ax2 + bx + c

Chemistry

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

a) k = 5.

b) remainder = -52.

Explanation:

When x = 2:

4a + 2b + c + 8 = 0...... (1)

When x = 3

9a + 3b + c + 27= 0........(2)

By the remainder theorem:

f(1) = -8 so

a + b + c + 1 = -8

a + b + c + 9 = 0...........(3)

Solving the system of equations:

equation (2) - equation (1) gives:

5a + b = -19

Equation (2) - (3) gives

8a + 2b = -18

Solving the last 2 gives:

a = -10 and b = 31.

So from the first equation:

c= -4a - 2b - 8

c = -4(-10) - 2(31) - 8 = -30

So the function is:

f(x) = x^3 - 10x^2 + 31x - 30

Now as the last term is -30 , because 2 roots are 2 and 3, making 2 factors (x-2) and (x- 3) the third factor is (x - 5), so the other root is 5

a) k = 5.

When the function is divided by x + 1 , (f-1) is the remainder.

b) f(-1) = (-1)^3 - 10(-1)^2 + 31(-1) - 30

= -1 + 10 - 31 - 30

= -52.

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il63 [147K]

Answer:

Here's what I get

Explanation:

1. Nickel sulfate

base + acid ⟶ salt + water

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Hydroxides of transition metals are insoluble; most sulfates are soluble.

$\underbrace{\hbox{Ni(OH)$_{2}$(s)}}_{\hbox{base}} + \underbrace{\hbox{H$_{2}$SO$_{4}$(aq)}}_{\hbox{acid}} \longrightarrow \, \underbrace{\hbox{NiSO$_{4}$(aq)}}_{\hbox{salt}} + \underbrace{\hbox{2H$_{2}$O(l)}}_{\hbox{water}}$

2. Carbonate + acid

Most carbonates are insoluble.

They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.

$\rm NiCO_{3}(s) + H_{2}SO_{4}(aq) \longrightarrow \, NiSO_{4}(aq) + H_{2}O(l) + CO_{2}(g)$

5 0

3 years ago

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$

Molar mass of A = $M_A$

The depression in freezing point of solution with B solute: $\Delta T_{f,B}$

Molar mass of B = $M_B$

$\Delta T_{f,A}>\Delta T_{f,B}$

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

$\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}$

$M_A$

This means compound B has greater molar mass than compound A,

4 0

3 years ago

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles$

b) moles of $I_2$

$\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles$

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

According to stoichiometry :

1 mole of $I_2$ require 1 mole of $H_2$

Thus 0.0787 moles of $l_2$ require= $\frac{1}{1}\times 0.0787=0.0787moles$ of $H_2$

Thus $l_2$ is the limiting reagent as it limits the formation of product and $H_2$ acts as the excess reagent. (10.0-0.0787)= 9.92 moles of $H_2$ are left unreacted.

Mass of $H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g$

Thus 19.9 g of $H_2$ remains unreacted.

5 0

3 years ago

In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n

artcher [175]

Stoichiomety:

1 moles of C + 1 mol of O2 = 1 mol of CO2

multiply each # of moles times the atomic molar mass of the compund to find the relation is weights

Atomic or molar weights:

C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
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Stoichiometry:

12 g of C react with 32 g of O2 to produce 44 g of CO2

Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen

And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.

You cannot obtain 72 g of CO2 from 18 g of C.

May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.

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4 years ago

Read 2 more answers

if you are told to get 100 mL of stock solution to use to prepare smaller size sample for an experiment, which piece of glasswar

fgiga [73]

Answer:

A beaker

Step-by-step explanation:

Specifically, I would use a 250 mL graduated beaker.

A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.

You don't need precisely 100 mL solution.

If the beaker is graduated, you can easily measure 100 mL of the stock solution.

Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).

6 0

3 years ago