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Dima020 [189]
3 years ago
5

Based on the kinetic theory, which statement is true?

Chemistry
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer:

(D) The particles of matter are arranged in different ways for the different states.

Explanation:

The particles of matter arrange differently based on the different states

They don't stay the same for each state. They have to change depending on certain things.

You might be interested in
What is the predicted change in the boiling point of water when 1.50 g of
dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

\textsf {While} \:  \sf  {\Delta T_b}  \: \textsf{expression is used} \\  \textsf {for elevation of boiling point}

⠀

⠀

<u>The</u><u> </u><u>elevation</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>phenomenon</u><u> </u><u>in</u><u> </u><u>which</u><u> </u><u>there</u><u> </u><u>is</u><u> </u><u>increase</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>in</u><u> </u><u>solution</u><u>,</u><u> </u><u>when</u><u> </u><u>the</u><u> </u><u>particular</u><u> </u><u>type</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>is</u><u> </u><u>added</u><u> </u><u>to</u><u> </u><u>pure</u><u> </u><u>solvent</u><u>.</u>

⠀

⠀

\sf  \large \underline{The \:  formula \: to \:  be  \: used \:  in \:  this \:  question \:  is}  \\   \boxed{T_b = i \times  K_b \times  m}

⠀

⠀

Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

⠀

'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

⠀

'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

⠀

⠀

<u>To</u><u> </u><u>find</u><u> </u><u>molality</u><u>,</u><u> </u><u>we</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>divide</u><u> </u><u>no</u><u>.</u><u> </u><u>of</u><u> </u><u>moles</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>by</u><u> </u><u>weight</u><u> </u><u>of</u><u> </u><u>solution</u>

⠀

While first we need to no. of moles

\sf \implies no. \: of \: moles =  \frac{weight \: of \: solute}{molar \: mass \: of \: solute}  \\  \\ \implies \sf no. \: of \: moles =  \frac{1.5}{208.23}  \\  \\  \sf \implies  no. \: of \: moles = 0.0072

⠀

⠀

<u>Now</u><u>,</u><u> </u><u>we</u><u> </u><u>will</u><u> </u><u>find</u><u> </u><u>molality</u>

⠀

\sf  \hookrightarrow molality =  \frac{no.\: of \: moles}{weight \: of \: solution}  \\  \\  \sf  \hookrightarrow molality =  \frac{0.072}{1.5}  \\  \\  \sf  \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}

⠀

⠀

\textsf{ \large{ \underline{Now substituting the required values}}}

⠀

\sf \longmapsto \Delta T_b = 3  \times 0.51  \times 0.0048 \\  \\ \\     \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}

⠀

⠀

⠀

<u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>change</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>7</u><u>3</u><u>5</u><u>°</u><u>C</u><u>.</u>

7 0
1 year ago
The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
3 years ago
How do minerals affect igneous rock formations?
wariber [46]
Mineral composition affects the classification of igneous rock. in simplified classification, igneous to is are classified by the type of feldspar present, by the I type a of feldspar present, or the absence of quartz. in case of neither present, then by the type of iron or magnesium present.
3 0
3 years ago
2. Extract the relevant information from the question: NaOH V = 30 mL , M = 0.10 M HCl V = 25.0 mL, M = ?
Andrei [34K]

Answer:

0.12M

Explanation:

A balanced equation for the reaction will go a great deal in obtaining our desired result. So, let us write a balanced equation for the reaction

HCl + NaOH —> NaCl + H2O

From the above equation,

nA (mole of the acid) = 1

nB (mole of the base) = 1

Data obtained from the question include:

Vb (volume of the base) = 30mL

Mb (Molarity of the base) = 0.1M

Va (volume of the acid) = 25mL

Ma (Molarity of the acid) =?

The molarity of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 25/ 0.1 x 30 = 1

Cross multiply to express in linear form

Ma x 25 = 0.1 x 30

Divide both side by 25

Ma = (0.1 x 30) / 25

Ma = 0.12M

The molarity of the acid is 0.12M

5 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
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