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Gelneren [198K]
3 years ago
10

Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. W

hat is the formula of the compound?
Chemistry
1 answer:
myrzilka [38]3 years ago
4 0

Answer:

RbI

Explanation:

The formula of a compound can be obtained on the basis of number of anion and cations in the unit cell. Iodide is present at the eight corners of the cubic cell.

So, no. of iodide ion = 1/8×8= 1

rubidium is present at the center

so, no. of rubidium ion = 1×1 =1

Therefore formula of the compound = RbI

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The volume occupied by 0.4g of hydrogen gas at S.T.P is ________
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4.48 dm^3

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As given in the problem, the conversion factor for the molar volume of all gases is 22.4 dm^3/mole at STP.  A mole of all ideal gases will occupy 22.4 liters (dm^3).  Calculate the moles of hydrogen contained in 0.4 grams of the gas by dividing the mass by the molar mass of H2.  H2 has a molar mass of 2 grams/mole.

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In a similar experiment, a current of 2.15 amps ran for 8 minutes and 24 seconds. The temperature of the water was 26.0°C. The v
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Answer:

\boxed{6.08 \times 10^{23}}

Explanation:

Data:

I = 2.15 A

t = 8 min 24 s

T = 26.0 °C

V = 65.4 mL

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1. Write the equation for the half-reaction

2H₂O ⟶ O₂ + 4H⁺ + 4e⁻

2. Calculate the moles of oxygen

p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}

3. Calculate the moles of electrons

\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}

4. Calculate the number of coulombs

t = 8 min 24 s =504 s

Q = It = 504 s × 2.10 C·s⁻¹= 1058 C

5. Calculate the number of electrons

\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}

6. Calculate Avogadro's number

N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}

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4 years ago
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