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Anna11 [10]
2 years ago
7

Which is true about the movement of gas particles? particles travel in straight-line paths until they collide with other objects

. the term "random walk" refers to the speed of particles. particles will partially fill their container. the motion of one particle depends on the motion of all the others. description?
Chemistry
2 answers:
Alenkasestr [34]2 years ago
5 0

<span>The answer is  ‘particles travel along straight-line paths until they collide with other objects.’. This is used to explain the Brownian movement of gas particles. As gas particles move rapidly, they collide with other molecules or the walls of a container, and change their direction randomly. It is impossible to predict the path of these particles after a collision hence ‘random walks’ phenomenon. </span>






sukhopar [10]2 years ago
5 0
<span>Answer: the first option: travel in straight-line paths until they collide with other objects.
</span><span />

<span>Justification:
</span><span />

<span>The kinetic molecular theory of gases deal with the movement of gas particles (molecules) to explain the behavior of the gases.
</span><span />

<span>These are the main assumptions made by the model:
</span><span />

<span>1) Size of the particles: The gases are formed by tiny particles whose volume is negligible in comparisson to the volume of the vessel that contain the gases.
</span><span />

<span>2) Intermolecular forces: There are not either attraction or repulsion forces between the molecules.


</span><span>3) Motion: the particles are in constant random motion. They move in straight line until they collide with each other or the walls of the vessel. The partilces do not lose energy when the collide with each other.</span>
<span /><span>
Those assumptions let to explain the low density of the gases, the compression and expansion, among other properties of the gases.

</span>
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What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon
Ulleksa [173]

Answer:

Theoretical yield = 31.8 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For Al_2O_3

Mass of Al_2O_3  = 60.0 g

Molar mass of Al_2O_3  = 101.96128 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{60.0\ g}{101.96128\ g/mol}

Moles_{Al_2O_3}= 0.5885\ mol

Given: For C

Given mass = 30.0 g

Molar mass of C  = 12.0107 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{30.0\ g}{12.0107\ g/mol}

Moles_{C}= 2.4978\ mol

According to the given reaction:

Al_2O_3+3C\rightarrow 2Al+3CO

1 mole of aluminium oxide react with 3 moles of carbon

0.5885 mole of aluminium oxide react with 3\times 0.5885 moles of carbon

Moles of carbon = 1.7655 moles

Available moles of carbon = 2.4978 moles

Limiting reagent is the one which is present in small amount. Thus, aluminium oxide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of aluminium oxide on reaction forms 2 moles of aluminium.

0.5885 mole of aluminium oxide on reaction forms 2\times 0.5885 moles of aluminium.

Moles of aluminium = 1.177 moles

Molar mass of aluminium = 26.981539 g/mol

Mass of sodium sulfate = Moles × Molar mass = 1.177 × 26.981539 g = 31.8 g

<u> Theoretical yield = 31.8 g</u>

3 0
2 years ago
Which missing item would complete this beta decay reaction?<br> PLATO
goldfiish [28.3K]

Answer: _{-1}^0\beta

Explanation:  

General representation of an element is given as:   _Z^A\textrm{X}

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Beta decay : In this process, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

The given reaction would be represented as:

_{43}^{98}\textrm{Tc}\rightarrow_{44}^{98}\textrm{Pb}+_{-1}^0\beta

7 0
2 years ago
Read 2 more answers
A graduated cylinder holds 100 mL of water. A lead weight is dropped into the cylinder bringing the new volume up to 450 mL. If
dolphi86 [110]

11.43g/mL

Explanation:

Given parameters:

Volume of water in the graduated cylinder = 100mL

Volume of water + lead weight = 450mL

Mass of lead weight = 4000g

Unknown:

Density of the lead weight = ?

Solution:

Density is the mass per unit volume of a body.

  Density  = \frac{mass}{volume}

Volume of the lead weight = volume of water displaced

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Density = \frac{4000}{350}  = 11.43g/mL

learn more:

Density brainly.com/question/2690299

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6 0
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Consider the nuclear equation below. Superscript 124 subscript 56 B a right arrow superscript 124 subscript 55 upper C s plus qu
nadezda [96]

Superscript o subscript negative 1 e.

Explanation:

The nuclear reactions is of 2 types, one is nuclear fusion and the other one is nuclear fission.

Nuclear fusion is nothing but the combining of 2 nuclei with an emission of energy along with an electron, proton or beta particle.

Nuclear fission is the break down of a nucleus into 2 or more nuclei along with an electron, proton or beta particle.

And the reaction is,

₅₆B¹²⁴  ₅₅C¹²⁴ +  ₋₁e⁰

So the blank was filled by means of a beta particle.

6 0
2 years ago
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Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
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