Malleability and density are examples of (C?)

Chemistry

2 answers:

sineoko [7]3 years ago

4 0

Answer: intensive properties because they do not depend on the amount of matter in the sample.

Explanation: Extensive properties are those properties which depend on the amount of the substance.

Intensive properties are those properties which do not depend on the amount of the substance.

Malleability and density are examples of intensive properties, as they do not depend on the amount of the substance.

I am Lyosha [343]3 years ago

3 0

If I'm not mistaken it's C.

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Choose the compound below that should have the highest melting point according to the ionic bonding model. A) AlN B) MgO C) NaF

iren2701 [21]

Answer:

The correct answer is C.

Explanation:

The compound with the highest melting point according to the ion bond model is the one with the greatest electronegativity difference, since the ion bond is stronger and therefore more difficult to break.

The electronegativity difference is calculated as the electronegativity difference of the atoms of the molecule.

AlN: Al (1,61) and N (3,04) so the difference gives 1,43.

MgO: Mg (1,31) and O (3,44) so the difference gives 2,13.

NaF: Na (0,93) and F (3,98) so the difference gives 3,05.

CaS: Ca (1) and S (2,58) so the difference is 1,58.

RbI: Rb (0.82) and I (2.66) so the difference is 1.84.

Analyzing the different cases, we can conclude that the molecule that presents the greatest difference in electronegativity is NaF, so it will have a higher boiling point due to the attraction force between its atoms.

Have a nice day!

5 0

3 years ago

Read 2 more answers

At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s2HIg=H2g+I2g Suppose a v

olya-2409 [2.1K]

Answer:

$\boxed{\text{28.0 s}}$

Explanation:

Whenever a question asks you, "How long does it take to reach a certain concentration?" or something like that, you must use the appropriate integrated rate law expression.

The integrated rate law for a first-order reaction is

$\ln \left (\dfrac{[A]_{0}}{[A]} \right ) = kt$

Data:

[A]₀ = 1.28 mol·L⁻¹

[A] = 0.17 [A]₀

k = 0.0632 s⁻¹

Calculation:

$\begin{array}{rcl}\ln \left (\dfrac{[A]_{0}}{0.170[A]_{0}} \right ) & = & 0.0632t\\\\\ln \left (5.882) & = & 0.0632t\\1.772 & = & 0.0632t\\\\t & = & \dfrac{1.772}{0.0632}\\\\t & = & \textbf{{28.0 s}}\\\end{array}\\\text{It will take } \boxed{\textbf{28.0 s}} \text{ for [HI] to decrease to 17.0 \% of its original value.}$