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r-ruslan [8.4K]
2 years ago
8

Data flows can be highly inconsistent, with periodic peaks, making data loads hard to manage. what is this feature of big data c

alled?
Chemistry
2 answers:
andrey2020 [161]2 years ago
8 0
Data flows can be highly consistent, with periodic peaks, making data loads hard to manage. This feature of big data is called variability.
kumpel [21]2 years ago
3 0

Data flow in Big Data is called Velocity or process speed.

Management of data that is structured in Big Data is called a Variety or types of data that can be accommodated.

<h2>Further explanation </h2>

Big data is a term that describes a large volume of data, both structured and unstructured that floods everyday business.

Big Data has three main characteristics:

  • Volume. The volume here is related to the size of the data storage media which is very large or may be unlimited. Big data has a very large amount of data so that the data processing requires large storage and more specific analysis is needed.
  • Velocity. Velocity here is a very fast speed at which data is received and (possibly) directly used. Usually, the highest speed of data flow is directly to memory compared to that written to disk. Some smart devices that use the internet operate in real-time or near real-time and will require real-time evaluation and action.
  • Variety. Variety in question here are the various types of data available. Traditional data types are usually more structured. With the development of big data, there is also unstructured data. Unstructured or semi-structured data such as text, audio, and video take time to process so you can know the meaning of these data.

Some things that are considered as big data are the following:

  • Internet use
  • Smartphone usage
  • Social media
  • Digitalization of media
  • Smart device

Here are some things that can be done with Big Data:

  • Customer relationship management (CRM)
  • Help your business operations become more efficient
  • Improve user mobile experience
  • Encourage innovation

There are five main steps to taking over this large "data structure" that includes traditional and structured data along with unstructured and structured data:

  • Set a big data strategy.
  • Identify the source of big data.
  • Access, manage and save data.
  • Data analysis.
  • Make decisions based on data.

Learn more

Big Data theory brainly.com/question/3521557

fields in Big Data brainly.com/question/3521557

Details

Class: high school

Subjects: Chemistry

Keywords: big data, analysis, data structure

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Which of the following chemical equations does not correspond to a standard molar enthalpy of formation? a. Mg(s) + C(s) + 3/2 O
goblinko [34]

Answer:

N₂_{s}  + O₂_{g}   →  2NO_{g}

Explanation:

  The given equations are:

      Mg_{s} + C_{s} + \frac{3}{2}O_{g}   →     MgCO₃_{s}

   

      C_{s} +  \frac{1}{2}O₂_{g}    →   CO_{g}

      N₂_{s}  + O₂_{g}   →  2NO_{g}

     

      N₂_{s}  + 2O₂_{g}    →  N₂O_{g}

      H₂_{g} + \frac{1}{2}O_{g}    →    HO_{l}

From the given equations only N₂_{s}  + O₂_{g}   →  2NO_{g} does not correspond to an expression of the standard molar enthalpy of formation.

  • The heat of formation is the heat liberated or absorbed when one mole of a compound is formed from its constituent elements under standard conditions.

 The equation must result in the formation of only one mole of the compound.

It could have been properly written as

   \frac{1}{2}N₂_{s}  +  \frac{1}{2}O₂_{g}   →  NO_{g}

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2 years ago
What impact would adding twice as much Na2 CO3 than required for stoichiometric quantities have on the quantity of product produ
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Answer:

There will be no observed impact of adding twice as much Na2CO3 on the product

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Stoichiometry gives the relationship between reactants and products in terms of mass, mole and volume.

If we consider the stoichiometry of the reaction, we will discover that the reaction occurs in a 1:1 ratio. This implies that use of twice the amount of Na2CO3 will only lead to an excess of Na2CO3 making the other reactant the limiting reactant. Once the other reactant is used up, the reaction quenches.

Hence, use of twice as much Na2CO3 has no impact on the quantity of product produced.

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Estimate the following: a) The volume occupied by 18 kg of ethylene at 55°C and 35 bar. b) The mass of ethylene contained in a 0
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Explanation:

(a)   The given data is as follows.

             T = 55^{o}C = (55 + 273) K

                = 328 K

             P = 35 bar = 3500 kPa

Let us assume that moles of ethylene present are 1 kmol and according to the ideal gas equation, PV = nRT.

Or,            V = \frac{nRT}{P}

                    = \frac{1 \times 8.314 \times 328}{3500 kPa}

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Hence, the volume occupied by 18 kg of ethylene at 55^{o}C and 35 bar is 0.779 m^{3}.

(b)   The given data is as follows.

              V = 0.25 m^{3}

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                P = 115 bar = 11500 kPa

Using ideal gas equation first, we will calculate its moles as follows.

            n = \frac{PV}{RT}

               = \frac{11500 \times 0.25}{8.314 \times 323 K}

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Since,  moles = \frac{mass}{\text{molar mass}}

Hence, mass of ethylene will be calculated as follows.

                moles = \frac{mass}{\text{molar mass}}

             1.07 mol = \frac{mass}{28 g/mol}

                   mass = 29.96 g

Therefore, mass of given ethylene is 29.96 g.

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2 years ago
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