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2 years ago

Avoltic cell prepared using aluminium and nickle has the flowing cell notation

Chemistry

1 answer:

SpyIntel [72]2 years ago

5 0

Answer:

Al (s) | Al³⁺ (aq) || Ni²⁺ (aq) | Ni (s)

Explanation:

It seems that you meant "A voltic cell".

Here, Al (s) represents the solid anode while Al³⁺ is in aqueous phase, these two forms are separated by a solid line. || is the salt bridge while Ni²⁺ is in aqueous phase and is reduced on the nickel cathode that is Ni (s).

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atroni [7]

The answer is D.) +0.34 V

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3 years ago

Which of the following statements best describes the number of neutrons in an atom?

Alexeev081 [22]

Answer:

its the last one

Explanation:

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3 years ago

Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1

bekas [8.4K]

Explanation:

$2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)$

Equilibrium constant of reaction = $K=154$

Concentration of NO = $[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M$

Concentration of bromine gas = $[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M$

Concentration of NOBr gas = $[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M$

The reaction quotient is given as:

$Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}$

$Q=328.06$

$Q>K$

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced. False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0

2 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.77 mol K₃N

<u>Step 2: Identify Conversions</u>

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

5 0

3 years ago

At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank

hodyreva [135]

Answer:

$\large \boxed{\text{-92 $^{\circ}$C}}$

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

$\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}$

2. Temperature in degrees Celsius

$\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}$

8 0

3 years ago

Avoltic cell prepared using aluminium and nickle has the flowing cell notation ​

Avoltic cell prepared using aluminium and nickle has the flowing cell notation