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PolarNik [594]
3 years ago
14

Three roots of a fifth degree polynomial function f(x) are –2, 2, and 4 + i. Which statement describes the number and nature of

all roots for this function?
Chemistry
2 answers:
Daniel [21]3 years ago
3 0
The two roots of this functions are the exact roots and one is an imaginary root. The -2 is the root of a negative number -16 because if you compute the -2 on its 3rd power you come up -16, the 2 is the root of 16 and the 4+i contains a imaginary number that is why it is an imaginary root
slava [35]3 years ago
3 0

Answer:

f(x) has three real roots and two imaginary roots.

Explanation:

It is given that the three roots of a fifth degree polynomial function f(x) are –2, 2, and 4 + i. In which -2,2 are real root and 4+i is imaginary root.

The number of roots of a polynomial is equal to the degree of that polynomial.

Since the degree of polynomial function is 5, therefore the function has total 5 roots.

According to the complex conjugate root theorem, if a+ib is a root of a polynomial, then its conjugate a-ib is also a root of that function. The number of imaginary roots are always an even number.

Since 4+i is a root of the polynomial, therefore 4-i is also a root of the function.

2 roots are real and 2 roots are imaginary. The remaining 1 root must be real because the number of imaginary roots can not be odd.

Therefore the function f(x) has three real roots and two imaginary roots.

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
17.8g of aluminum sulfate is produced from how many grams of potassium sulfate?​
german

Answer:

1 mole of ferric contains 2 moles of iron,and 12 moles of oxygen atoms, and three moles of sulphate ions

3 0
3 years ago
Millions of years ago in many different regions of the world, large-scale burials of organic matter were followed by the formati
kramer
The answer is A: Areas where the geologic process occurred now have major petroleum reserves
5 0
2 years ago
Read 2 more answers
If 2.03 g of oxygen react with carbon monoxide, how many grams of CO2 are formed?
mamaluj [8]

Answer:

The suitable equation for this reaction is

2CO + O₂ -----> 2CO₂

Here, we are given that we have 2 grams of O₂

From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂

Moles of O₂:

2/32 = 1/16 moles

Therefore, the number of moles of CO₂ is twice the moles of O₂

Moles of CO₂ = 2 * 1/16

Moles of CO₂  formed = 1/8 moles

Mass of CO₂  formed = Molar mass of CO₂ * Moles of CO₂

Mass of CO₂  formed = 44 * 1/8

Mass of CO₂  formed = 5.5 grams

Hence, option B is correct

Kindly Mark Brainliest, Thanks!!!

8 0
3 years ago
the concentration of the radio active isotope potassium-40 in a rock sample is found to be 6.25%. what is the age of the rock
julsineya [31]

Answer:

5.0 x 10⁹ years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • The half-life of K-40 = 1.251 × 10⁹ years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (K-40) ([A₀] = 100%).

[A] is the remaining concentration of (K-40) ([A] = 6.25%).

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.

∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.

8 0
3 years ago
Read 2 more answers
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