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sergeinik [125]

3 years ago

11

(EASY!!) What is the name for this structure?

2 answers:

nikklg [1K]3 years ago

6 0

Answer:

ccc

Explanation:

yeah

Fantom [35]3 years ago

5 0

Answer:

uhh I can't really tell- it looks like a lot of things

Explanation:

it kind of looks like a trapezoid or maybe a "gun" I haven't really seen a shape of this kind- so I'm going to say either one of those answers. sorry if I wasn't much help (◞‸◟ㆀ)

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. Metallic iron has a body-centered cubic lattice with all atoms at lattice points and a unit cell whose edge length is 286.6 pm

adell [148]

Answer:

$\large \boxed{\text{55.8 u}}$

Explanation:

1. Calculate the volume of the unit cell

V = l³ = (2.866 × 10⁻⁸ cm)³ = 2.354 × 10⁻²³ cm³

2. Calculate the mass of a unit cell

$\text{Mass} = 2.866 \times 10^{-23}\text{ cm}^{3} \times \dfrac{\text{7.87 g}}{\text{1 cm}^{3}} = 1.853 \times 10^{-22} \text{ g}$

3. Calculate the mass of one atom

A body-centred unit cell contains two atoms.

$\text{Mass of 1 atom} = \dfrac{1.853 \times 10^{-22} \text{ g}}{\text{2 atoms}} \times \dfrac{\text{1 u}}{1.661 \times 10^{-24}\text{ g}} = \textbf{55.8 u}\\\\\text{The molar mass of Fe from the Periodic Table is $\large \boxed{\textbf{55.845 g/mol}}$}$

6 0

3 years ago

You are given the following three half-reactions:(1) Fe³⁺(aq) + e⁻ ⇄ Fe²⁺(aq) (2) Fe²⁺(aq) +2e⁻ ⇄ Fe(s) (3) Fe³⁺(aq) +3e⁻ ⇄ Fe(s

RideAnS [48]

The E°half-cell of the (3) is 0.33 volts .

Given,

(1) Fe3+(aq) +e =Fe2+(aq)

(2) Fe2+(aq) +2e = Fe(s)

(3) Fe3+(aq) +3e = Fe(s)

We know ,

E° half-cell of (1) is 0.77 volts .

E° half cell of ( 2) is -0.44 volts .

By resolving or adding equation 1 and 2 we will get (3) ,

Thus , the value of the E° half cell of (3) is given by ,

E°cell = E° half-cell of (1 ) + E° half-cell of (2 ) = 0.77-0.44 = 0.33 volts

Hence the E° half-cell of ( 3 ) is 0.33 volts .

<h3>What is cell potential? </h3>

It is the difference between the electrode potentials of two half cells.

it also defined as the force which causes the flow of electrons from one electrode to the another and thus results in the flow of current.

Ecell = E( cathode) - E ( anode)

<h3>What is electrode potential ? </h3>

The tendency of the electrode to lose of gain electrons is called as electrode potential.

Learn more about cell potential here :

brainly.com/question/19036092

#SPJ4

4 0

1 year ago

PLEASE HELP ASAP

Leni [432]

Answer:

the answer is B

Explanation:

the answer is B

4 0

3 years ago

Read 2 more answers

If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?

vlada-n [284]

Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
Cross-multiply: $(5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)$
Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate <em>x</em>: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em /> $18.0403 \ L \ H_2 \approx 18 \ L \ H_2$ <em />

5 0

3 years ago

Help pls! thank u (:

alexgriva [62]

Answer:

7.65 moles of silver are produced

Explanation:

Zinc, Zn, reacts with silver nitrate, AgNO3, as follows:

Zn + 2AgNO3 → Zn(NO3)2 + 2Ag

<em>Where 1 mole of Zn reacts with an excess of AgNO3 to produce 2 moles of Ag</em>

To solve this question we must convert the mass of Zn to moles and, using the chemical equation, we can find the moles of Ag as follows:

<em>Moles Zn (Molar mass: 65.38g/mol):</em>

250g Zn * (1mol / 65.38g) = 3.824 moles Zn

<em>Moles Ag:</em>

3.824 moles Zn * (2mol Ag / 1mol Zn) =

<h3>7.65 moles of silver are produced</h3>

7 0

3 years ago