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1 year ago

What is the coordination number for atoms that pack within the diamond cubic crystal structure?.

1 answer:

6 0

Coordination numbers are defined for the crystalline structures that tell about the numbers of atoms bonded to the central atom. The coordination number of a diamond is 4.

<h3>What is a coordination number?</h3>

A coordination number has been defining the number of the ions, atoms, and molecules that are bonded to the central atom of the crystalline compound or a complex.

The diamond has 8 carbon atoms in its one unit and its coordination number is 4 as its lattice type is face-centered cubic that has carbons at the corners and face center of the structure. One carbon contributes 1/8 hence and the faces center contributes 1/2.

Therefore, the diamond has a coordination number of 4.

Learn more about coordination numbers, here:

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Grace runs 6miles in 2 hours what is her speed??

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Speed = miles / time

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3 years ago

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Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
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Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

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2 years ago

The accepted value for the molar volume of a gas is 22.4 L. Sarah's experimental data indicated that a mole of a gas had a volum

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Answer:

Percent error = 12.5%

Explanation:

In a measurement you can find percent error following the formula:

Percent error = |Measured value - Accepted Value| / Acepted value * 100

Based on the data of the problem, accepted value is 22.4L and the measured Value (Value of Sara) was 19.6L.

Replacing:

Percent error = |Measured value - Accepted Value| / Acepted value * 100

Percent error = |19.6L - 22.4L| / 22.4L * 100

Percent error = |-2.8L| / 22.4L * 100

Percent error = 2.8L / 22.4L * 100

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2 years ago

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Explanation:

The two ways that energy can be transferred are by doing work and by heat transfer.

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2 years ago

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

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3 years ago