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3 years ago

Please help I just need those last 4! :)

Chemistry

1 answer:

mojhsa [17]3 years ago

5 0

Answer:

Series:

3. Thermistor

Parallel:

1. ???

2. ???

3. ???

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PLEASE HELP ME :( It’s urgent!!!!!

MrRa [10]

Answer:

C and D

Explanation:

In the first table, both element particles have more electrons than protons which is unusual and indicates they have a negative charge. Furthermore both numbers add up to full electron shells (2,8 and 2,8,8) meaning they are ions (charged particles formed to gain full valence shells and become stable), so it is C for the first question, negative ions.

As for the second one, isotopes like versions of the same element except with different numbers of neutrons, so they are still the same element but have different mass. As shown here the numbers indicate the number of neutrons therefore it is D, 3 more neutrons and no more electrons.

Hope this helped!

5 0

3 years ago

Initial Volume (VI) Final Volume (Vf) = Object's volume =

Blizzard [7]

Final volume minus the initial equals objects volume

3 0

3 years ago

What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of th

zaharov [31]

Answer:

0.35 V

Explanation:

(a) Standard reduction potentials

E°/V

Fe²⁺ + 2e- ⇌ Fe; -0.41

Cr³⁺ + 3e⁻ ⇌ Cr; -0.74

(b) Standard cell potential

E°/V

2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74

3Fe ⇌ 3Fe²⁺ + 6e-; -0.41

2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33

3. Cell potential

2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr

3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-

2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)

The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF}\ln Q$

(a) Data

E° = 0.33 V

R = 8.314 J·K⁻¹mol⁻¹

T = 298 K

z = 6

F = 96 485 C/mol

(b) Calculations:

$Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}$