Question 1 of 10

What is technology?

A. An understanding of something new.

B. The steps that engineers go through to create a product.

C. Something created using science for use by society.

D. A method that is used to solve problems,

SUBMIT

Answers

Answer 1

Answer:

C.

Explanation:

You can use the series of elimination for this. First, you look at A. Technology is not an understanding of something new, so we cross that out. Second, you look at B. Technology isn't a series of steps, so we can mark that one off. Third, you look at C. Technology is something created using science for use by society, so we can keep that in mind. Lastly, we check D. Technology doesn't match up to the definition, so we can cross that one out. The answer that would make the most sense would be D.


Related Questions

Why does sodium chloride form a crystal lattice

Answers

Answer:

Ions of opposite charge strongly attract each other; those of like charges repel. As a result ions in an ionic compound are arranged in a particular manner.

Explanation:

Google is smart

PLEASE HELP! WILL DO BRAINLIEST! What do scientists call all of the compounds that contain carbon and are found in living things?
organic

inorganic

acidic

nonacidic

Answers

Answer:

acidic because of electrical issues and the body of electrical equipment

If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction

Answers

Answer:

7.71 × 10⁻⁴ M/s

Explanation:

The initial rate of the reaction can be expressed by using the formula:

[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]

where;

Pressue P = 1.00 atm

Volume V =5.74mL =  (5.74 /1000) L

Rate R = 0.082 L atm/mol.K

Temperature = 298 K

[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]

= 2.35 × 10⁻⁴ mol

Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]

Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]

Δ[O₂]  = 0.04626 M

The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

= [tex]\dfrac{0.04626}{60}[/tex]

= 7.71 × 10⁻⁴ M/s

What two types of elements make up an Ionic bond ?

Answers

Answer:

Metals and Non-metals

Explanation:

What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.

Answers

Answer:

i

[tex]J_{m} = 20 [/tex]

ii

[tex]J_{m} = 22.5 [/tex]

Explanation:

From the question we are told that

  The first temperatures is [tex]T_1 =  25^oC =  25 +273 =298 \ K[/tex]

   The second temperature is  [tex]T_2 =  100^oC =  100 +273 = 373 \ K[/tex]

Generally the equation for  the most highly populated rotational energy level is mathematically represented as

     [tex]J_{m} = [ \frac{RT}{2B}]  ^{\frac{1}{2} } - \frac{1}{2}[/tex]

Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]

Also  

      B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]

   Generally the energy require per mole to move 1 cm is  12 J /mole

So   [tex]0.244 \ cm^{-1}[/tex]  will require x J/mole

           [tex]x =  0.244 *  12[/tex]

=>          [tex]x =  2.928 \ J/mol [/tex]

So at the first temperature

     [tex]J_{m} = [ \frac{8.314 * 298  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 20 [/tex]

So at the second temperature

           [tex]J_{m} = [ \frac{8.314 * 373  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 22.5 [/tex]

If the earth was a guava fruit, the space where the seeds are would be the core/mantle​

Answers

Right on ! I need to answer a question to get mine answered so here I am :)

When you finish exercising, you are hot, tired, and sweating. After a bottle of juice, you feel a lot better. Which organ systems are working together in this scenario?

Answers

Answer:

Nervous and Excretory

Explanation:

The nervous system makes you thirsty. The integumentary system makes you sweat. the integumentary system is very similar to the Excretory system.

Nervous and Excretory systems are working together in this scenario.

The nervous system makes you thirsty. The integumentary system makes you sweat. the integumentary system is very similar to the Excretory system.

The blood flow increases, your brain is exposed to more oxygen and nutrients. Exercise also induces the release of beneficial proteins in the brain. Sweating cools the body when it becomes warm. When the body temperature rises, such as when exercising on a hot day, the dermal blood vessels dilate.The excretory system works with the endocrine system to help maintain homeostasis.

Learn more:

brainly.com/question/17342396

Solid diarsenic trioxide reacts with fluorine gas (F2) to produce liquid arsenic pentafluoride and oxygen gas (O2). Write the Qc for this reaction.

Answers

Answer:

QC= [O2]^3/[F2]^10

Explanation:

In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.

Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?

Answers

Answer:

a) - 0.2 M

b) - 0.2 M

c)- 0

Explanation:

The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:

MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol

a). Molarity = moles CuBr₂/1 L solution

moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol

Volume in L = 375 mL x 1 L/1000 mL = 0.375 L

M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M

b). When is added to water, CuBr₂ dissociates into ions as follows:

CuBr₂ ⇒ Cu²⁺ + 2 Br⁻

We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:

0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M

c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).

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