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3 years ago

In a metallic bond electrons are _______________. A. transferred C. shared B. delocalized D. lost

Chemistry

1 answer:

sergiy2304 [10]3 years ago

4 0

Answer:

in a metallic bond electrons are delocalized

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Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat

Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

Density of aluminum (ρ): 2.7 g/cm³

Mass of aluminum (m): 81 g

Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

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Ymorist [56]

Plastic packaging requires oil to create it

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3 years ago

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P

expeople1 [14]

Answer : The ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

Solution : Given,

$K_p=1.5\times 10^6$

$p_{O_2}$ = 0.21 atm

The given equilibrium reaction is,

$NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}$

Now put all the given values in this expression, we get:

$1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}$

$\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}$

$\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5$

Therefore, the ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

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3 years ago

Based on the simulation activity and understanding how magnets have their own magnetic field explain what might make the magneti

Scrat [10]

Answer:

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Explanation:

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BabaBlast [244]

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