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3 years ago

What are the solutions to the quadratic equation 2x2 + 10x - 48 = 0?

Chemistry

1 answer:

Natali [406]3 years ago

8 0

Answer:

Explanation:

$2x^2+10x-48\\=2\left(x^2+5x-24\right)\\x^2+5x-24\\=\left(x^2-3x\right)+\left(8x-24\right)\\=x\left(x-3\right)+8\left(x-3\right)\\=\left(x-3\right)\left(x+8\right)\\=2\left(x-3\right)\left(x+8\right)\\2\left(x-3\right)\left(x+8\right)=0\\x-3=0\\x = 0+3\\x = 3\\x+8=0\\x+8-8=0-8\\x=-8\\x=3,\:x=-8$

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ANSWER ASAP thanks

weqwewe [10]

Answer:

Option D: it's ability to lose electrons

Explanation:

Alkali metals are usually discovered in nature. They have highly reactivity at STP conditions (standard temperature and pressure conditions) and easily lose their outermost electron to form positive ions known that have a charge of +1.

Thus, what can determine the extent of reactivity of an alkali metal, is it's ability to lose electrons

5 0

3 years ago

Read 2 more answers

1. What % of the nation's energy is used by the Industrial<br> Sector?

kari74 [83]

Answer:

Petroleum:92 Percent

Natural Gas:3 Percent

Renewable energy:5 Percent

Explanation:

US primary energy consumption by source and sector (2017)[17]

Supply sources Percent of source Demand sectors Percent of sector

Petroleum

36.2% 72% Transportation

23% Industrial

5% Residential and commercial

1% Electric power Transportation

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3% Natural gas

5% Renewable energy

Natural gas

28.0% 3% Transportation

35% Industrial

28% Residential and commercial

34% Electric power Industrial

21.9% 38% Petroleum

45% Natural gas

5% Coal

12% Renewable energy

Coal

13.9% 9% Industrial

<1% Residential and commercial

91% Electric power Residential and commercial

10.4% 16% Petroleum

76% Natural gas

<1% Coal

8% Renewable energy

Renewable energy

11.0% 13% Transportation

23% Industrial

7% Residential and commercial

57% Electric power Electric power

37.2% 1% Petroleum

26% Natural gas

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23% Nuclear electric power

Nuclear electric power

8.4% 100% Electric power

3 0

3 years ago

Chloroform is an excellent solvent for extracting caffeine from water. The distribution coefficient, KD, (Cchloroform/Cwater) fo

Art [367]

The relative volumes of chloroform and water that should be used is 9:10

Concentration of solution in chloroform = $90$ ( moles of chloroform )

Concentration of solution in water = $10$ ( moles of water )

Dissociation constant at $25^oC$ ; $K_D = 10$

$K_D =$ Concentration of solution in chloroform / Concentration of solution in water

Meaning;

$K_D = \frac{\frac{mole\ of\ chloroform}{volume\ of\ chloroform} }{\frac{mole\ of\ water}{volume\ of\ water} }$

Since $90$ mole is present in chloroform and $10$ mole is present in water, Total mole of Caffeine present is $100$

Now, we substitute our given values into the equation

$10 = \frac{\frac{90}{volume\ of\ chloroform} }{\frac{10}{volume\ of\ water} }\\\\10 *\frac{10}{volume\ of\ water} = \frac{90}{volume\ of\ chloroform} \\\\\frac{100}{volume\ of\ water} = \frac{90}{volume\ of\ chloroform}\\\\\frac{volume\ of\ chloroform}{volume\ of\ water} = \frac{90}{100}\\\\ \frac{volume\ of\ chloroform}{volume\ of\ water} = \frac{9}{10}\\\\ \frac{volume\ of\ chloroform}{volume\ of\ water} = 9:10$

Therefore, the relative volumes of chloroform and water that should be used is 9:10

Learn more; brainly.com/question/11060225

8 0

3 years ago

Write the balanced NET IONIC equation for the reaction that occurs when ammonium nitrate and potassium hydroxide are combined. N

vova2212 [387]

Answer:

Net Ionic equation

NH₄⁺ + OH⁻ → NH₃ + H₂O

Option B is correct.

Weak Acid Strong Base

Check Explanation for the extent of the reaction.

Explanation:

Ammonium nitrate = NH₄NO₃

Potassium Hydroxide = KOH

Ammonium salts combine with alkalis to liberate NH₃ and form water.

The two reactants combine to give

NH₄NO₃ + KOH → KNO₃ + NH₃ + H₂O

In ionic form,

- NH₄NO₃ exists as NH₄⁺ and NO₃⁻

- KOH exists as K⁺ and OH⁻

- KNO₃ as K⁺ and NO₃⁻

And NH₃ and H₂O stay as they are, as per covalent compounds.

So, we have

NH₄⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + NH₃ + H₂O

Eliminating the ions that exist on both sides, we have the net ionic equation to be

NH₄⁺ + OH⁻ → NH₃ + H₂O

which shows that this reaction is essentially a neutralization reaction in which the Bronsted Lowry acid, NH₄⁺, loses its proton to the base, OH⁻ and gives conjugate base, NH₃ and conjugate acid, H₂O.

This reaction is classified as a Weak acid versus Strong Base reaction as NH₄⁺ is from a Weak acid and OH⁻ is from a strong base.

Since this reaction is between a Weak base and a strong acid, the ionization isn't expected to be 100%, Hence, the extent of this reaction will be any option that is not 100%, a couple pieces of information might be required for the correct estimate, but above 50% seems correct.

Hope this Helps!!!

8 0

3 years ago

Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i

anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

$K=\frac{Products}{reagents}$

In any case

aA +Bb equilibrium Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so $K_{2}$ will be less than $K_{1}$ the forward reactions is not favored.

${e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

5 0

3 years ago

What are the solutions to the quadratic equation 2x2 + 10x - 48 = 0?​

What are the solutions to the quadratic equation 2x2 + 10x - 48 = 0?