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3 years ago

Which molecules would most likely cause a liquid to have the lowest viscosity?

2 answers:

8 0

The molecules with the lowest viscosity is the liquid with the least complex molecular structure. The structure affects the viscosity of the fluid because they are less compact and can flow freely.

valina [46]3 years ago

4 0

Explanation:

Hello! Let's solve this!

Viscosity is a physical property of fluids.

The factors that affect viscosity are: temperature, gas and pressure. In the case of molecules that can affect the lower viscosity, they are the ones with a simpler molecular structure.

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Consider the dissolution of ammonium nitrate. 1.25 g of amminum nitrate is dissolved in enough water to make 25.0 mL of solution

Mrac [35]

Answer:

$\Delta H=-0.02\ kJ$

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$ is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 1.25 g

Specific heat = 4.18 J/g°C

$\Delta T=21.9-25.8\ ^0C=-3.9\ ^0C$

So,

$\Delta H=-1.25\times 4.18\times 3.9\ J=-20.3775\ J$

Negative sign signifies loss of heat.

Also, 1 J = 0.001 kJ

So,

$\Delta H=-0.02\ kJ$

5 0

3 years ago

Calculate the cell potential, the equilibrium constant, and the free energy change for

Tresset [83]

Answer:

Explanation:

Ba(s) + Mn²⁺ (aq,1M) → Ba²⁺ (aq,1M) + Mn(s)

Ba⁺²(aq) +2e → Ba(s) , E° = −2.90 V

Mn⁺²(aq) +2e → Mn(s), E⁰ =0.80 V

Anode reaction :

Ba(s) → Ba⁺²(aq) +2e E° = −2.90 V

Cathode reaction :

Mn⁺²(aq) +2e → Mn(s) E⁰ =0.80 V

Cell potential = Ecathode - Eanode

Ecell = .80 - ( - 2.90 )

Ecell = 3.7 V .

equilibrium constant ( K ) :

Ecell = .059 log K / n

n = 2

3.7 = .059 log K / 2

log K = 125.42

K = 2.63 x 10¹²⁵ .

Free energy change :

ΔG = - n F Ecell

= - 2 x 96500 x 3.7

= 714100 J

= 7.141 x 10⁵ J .

5 0

3 years ago

A____ and _______ takes the shape of their containers

MissTica

Answer:

liquid

gas

Explanation:

liquid has an indefinite shape

gas has an indefinite shape

3 0

2 years ago

Read 2 more answers

Do the elements in group 7 of the periodic table react with metals to form covalent compounds, ionic compounds, neither, or both

Vladimir [108]

Answer:Group 1 elements have 1 valence electron, meaning they have 1 extra electron that can easily be donated to an atom in search of 1 more electron. When they give away that extra electron to form an ionic compound, they become more stable.

For example, Group 7A (Group 17) elements have 7 valence electrons, meaning they need 1 extra electron to be stable. Group 1 and 7A elements make splendid ionic compounds.

Explanation:

7 0

2 years ago

In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,

Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - -

At t =equilibrium (x-s) s 2s

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$K_{sp}=s\times {2s}^2$

$K_{sp}=4s^3$

Given s = 7.4×10⁻³ M

So, Ksp is:

$K_{sp}=4\times (7.4\times 10^{-3})^3$

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

CuF₂ ⇄ Cu²⁺ + 2F⁻

At t=0 x - 0.20

At t =equilibrium (x-s') s' 0.20+2s'

The expression for Solubility product for CuF₂ is:

$K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2$

$1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2$

Solving for s', we get

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>

3 0

3 years ago