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Alenkasestr [34]
3 years ago
8

Pls help me I need this answer ______ are easier to work with than reality and can be manipulated. Computers Observations Experi

ments Models
it is not observations
Chemistry
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

Models

Explanation:

Models are abstractions of the real world. Models helps to mimick what is actually the true representation of a body. It is easier to work with models because they are our own creation. This makes them flexible to carryout any form of adjustments and manipulations. We can try out a model with different variables to simulate an obtainable result.

Models are easier to work with than the complex reality with a lot of factors and conditions that may not be easily predictable and under control.

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Talja [164]

yes

that is truly epic if you understand

4 0
4 years ago
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0
3 years ago
Pls help
navik [9.2K]
The Cambrian period, part of the Paleozoic era, produced the most intense burst of evolution ever known. The Cambrian Explosion saw an incredible diversity of life emerge, including many major animal groups alive today. Among them were the chordates, to which vertebrates (animals with backbones) such as humans belong. These included brachiopods, which lived in shells resembling those of clams or cockles, and animals with jointed, external skeletons known as arthropods—the ancestors of insects, spiders, and crustaceans. These toughened-up creatures represented a crucial innovation: hard bodies offering animals both a defense against enemies and a framework for supporting bigger body sizes.
3 0
3 years ago
Which rate law is consistent with the mechanism?
yulyashka [42]
C I’m pretty sure. The first step is the rate determining step
6 0
3 years ago
Consider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X.The enthalpy of formation of MX is ΔH
Studentka2010 [4]
Here are the given:
<span>ΔHf° = –423 kJ/mol
</span> ΔHsub = 119 kJ/mol
IE = 469 kJ/mol
ΔHEA = –301 kJ/mol
BE = 161 kJ/mol

The lattice energy of the compound is solved using the formula:
U = <span>ΔHf° - </span>ΔHsub - BE - IE - ΔHEA
U = -423 - 119 - 161 - 469 - (-301)
U = -871 kJ/mol

Therefore, the lattice energy is 871 kJ/mol (released).
7 0
3 years ago
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