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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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Vesnalui [34]

Answer:

well there are many ways to prevent a pest infestation ..... as we all know ... insects such as ants love leftovers, so store food in airtight containers like Tupperware or jars with rubber seals on the lids.

Keep a tight lid on your trash can at all times and move the trash to an outside dumpster as soon as possible.

Clean your counter tops, tables, and floors on a regular basis to prevent crumbs or food buildup

Explanation:

Hope this helps:)

3 0
3 years ago
Which pH corresponds with the acidic quality of the rainwater?<br> a.6<br> b.7<br> c.8<br> d.9
jeka94
The lower you go, the more acidic. The higher you go, the more alkaline. Your answer would most likely be 6.
8 0
3 years ago
1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to
Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0
2 years ago
Care to help me guys please​
earnstyle [38]

Answer:

A. Na₂SO₄ and HCl

C. Polar solutes are soluble in polar solvents but are insoluble in non-polar solvents Non-polar solutes are insoluble in polar solvents but are are soluble in non-polar solvents

Ionic solutes are soluble in polar solvents but are insoluble in non-polar solvents.

Like dissolves like simply means that molecules of substances having similar chemical properties dissolve in each other

Explanation:

A. Ionic substances like Na₂SO₄ are composed of charged particles called ions. These ions are either positively charged or negatively charged, therefore, they are attracted to substances of opposite charges. Also, polar molecules like HCl contains two oppositely charged ends. A polar solvent consists of molecules with two oppositely charged ends, therefore, ionic substances as well polar substances dissolve in them according to the concept of like dissolves like.

Gasoline being non-polar will only dissolve in like substances, polar solvents.

C. Polar solutes are soluble in polar solvents but are insoluble in non-polar solvents Non-polar solutes are insoluble in polar solvents but are are soluble in non-polar solvents

Ionic solutes are soluble in polar solvents but are insoluble in non-polar solvents.

The statement "Like dissolves like" simply means that molecules of substances having similar chemical properties dissolve in each other. For example gasoline, a non-polar substance will dissolve only in a non-polar solvent like kerosene. Also, HCl, a polar molecule will dissolve in a polar solvent like water.

7 0
2 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
2 years ago
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