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3 years ago

An engineer uses aluminum to build an airplane rather than composite

Chemistry

1 answer:

lesantik [10]3 years ago

5 0

Answer:

Explanation:

I would assume A because it says in the problem statement that chooses aluminum because he is worried that composite will suddenly fail... therefore he would use aluminum because it would bend before breaking and therefore show warning.

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Some of the physical properties used to distinguish between the three groups are: _________

Pavlova-9 [17]

The Group 3A metals have three valence electrons in their highest-energy orbitals (ns2p1). They have higher ionization energies than the Group 1A and 2A elements, and are ionized to form a 3+ charges. The Group 3A metals are silvery in appearance, and like all metals are good conductors of electricity.

7 0

3 years ago

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How are ironic bonds and covalent bonds different and how are they different?

Vinil7 [7]

Ionic bonds are forces that hold together electrostatic forces of attractions between oppositely charged ions. Ionic bonds have an electronegativity difference greater than or equal to 2. Covalent bonds have an electronegativity difference that is less than 2.

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3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

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3 years ago

To name the compound written as CuCl2, you would write:

vovangra [49]

Answer: option C. Copper (II) chloride

Explanation:

To name CuCl2, we need to know the oxidation state of Cu in the compound as chlorine always have oxidation on —1 in all its compound. The oxidation state of Cu can be calculated as follows:

Cu + 2Cl = 0 (since the compound has no charge)

Cl = —1

Cu + 2(—1) = 0

Cu —2 = 0

Collect like terms

Cu = 0 +2

Cu = +2

Therefore, the oxidation state of Cu in CuCl2 is +2.

The name of the compound will be copper(ii) chloride, since cupper has oxidation state +2 in the compound.

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2 years ago

Balance the following chemical equation <br> __Mg+__O2–>__MgO

mario62 [17]

Answer:

2Mg^+ +O2 right arrow 2MgO

Explanation:

6 0

3 years ago

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