All categories

2 years ago

A student has a piece of aluminum metal. Which is the most reasonable assumption the student could make about the metal?

Chemistry

2 answers:

alukav5142 [94]2 years ago

4 0

B. It could melt and boil if thrown into a campfire.

"In fire tests on aluminium materials, when the temperature exceeds the melting point, in the range 600-660°C, the aluminium surface exposed to the fire can be seen to melt, but it does not burn."(www.alfed.org.uk/files/Fact%20sheets/11-aluminium-and-fire.pdf)

Rudik [331]2 years ago

3 0

The metals are elements which have low ionization potential and can lose electrons easily

the other physical characteristics associated with them and hence with Aluminium are

a) they can conduct electricity in molten state, hence It would only conduct electricity if it were melted.

b) It could be stretched into a thin wire. It means it is ductile.

In case of given aluminium metal the correct answer is

it could be stretched to thin wire

You might be interested in

A common way of initiating certain chemical reactions with light involves the generation of free halogen atoms in solution. if δ

STatiana [176]

Answer: The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

Explanation:

$Cl_2\overset{h\nu}\rightarrow Cl^-,Delta H_{rxn}=242.8kJ/mol$

Energy required to produce free chlorine atoms from one mole of chlorine gas :

= 242.8kJ = $242.8\times 1000=242800 Joules$ (1kJ=1000J)

1 mole = $6.022\times 10^{23}$ molecules

For $6.022\times 10^{23}$ molecules = 242,800 Joules

For one molecule of chlorine gas = $\frac{242800 Joules/mol}{6.022\times 10^{23} mol^{-1}}=40,318.83\times 10^{-23}Joules$

According to photoelectric equation:

$E=h\nu=\frac{hc}{\Lambda }$

E = Energy of the photon of light used to produce free chlorine atoms

$\nu$ = frequency of the light used to produce free chlorine atoms

h = Planck's constant = $6.626\times 10^{-34}J.s$ , c = speed of light= $3\times 10^8 m/s$

$\lambda$ = wavelength of the light used to produce free chlorine atoms

$40,318.83\times 10^{-23}J=\frac{hc}{\Lambda }=\frac{6.626\times 10^{-34} J.s\times 3\times 10^8 m/s}{\lambda }$

$\lambda=0.0004930203\times 10^{-3} m=493.0203\times 10^{-9} m=493 nm$

The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

5 0

2 years ago

Two pure elements react to form a compound. One element is an alkali metal, X, and the other element is a halogen, Z. Which of t

Lelechka [254]

Answer:

Option c. It contians ionic bonds

Explanation:

Let's take as an example NaCl

An element from group 1 (alkali) with an halogen.

Elements from group 1 form cations with +1 (charge)

This ionic salt, can't never has the formula XZ₂

Ionic salts, also can be dissolved in water, and they don't contain covalent bonds.

4 0

2 years ago

Read 2 more answers

If angle ABE = 2n + 7 and angle EBF=4n-13,<br>find angle ABE.

andreev551 [17]

Answer:

The measure of angle ABE is 27°.

Explanation:

Consider the below figure attached with this question.

Given information: ∠ABE=2n+7 and ∠EBF=4n-13.

From the below figure it is clear that ∠ABE and ∠EBF are congruent.

$m\angle ABE=m\angle EBF$

$2n+7=4n-13$

Isolate variable terms on one side.

$7+13=4n-2n$

$20=2n$

Divide both sides by 2.

$10=n$

The value of n is 10.

We need to find the measure of ∠ABE.

$\angle ABE=2(10)+7=20+7=27$

Therefore, the measure of angle ABE is 27°.

4 0

2 years ago

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 10.0 gg of aluminum? Express your answ

gayaneshka [121]

Answer:

The answer is 0.370moles (3 s.f)

Explanation:

Step 1: write the balanced equation for the reaction

Equation for the reaction:

2Al(s) + 6HCl(g) ---------> 2AlCl3(g) + 3H2(g)

Step 2: Equate the mole of the needed substances

So therefore,

2moles of aluminum will produce 2 moles of aluminum chloride

Note that:

Relative atomic mass of Al = 27 g/mole

And, 27g of Al = 1 mole of Al.

Step 3: Solve for the required number of mole.

2moles of Al = 2moles of AlCl3

Same as

1 mole of Al = 1 mole of AlCl3

27g of Al = 1 mole of AlCl3

10g of Al = (1/27 * 10g) of AlCl3

10g of Al = 0.370moles of AlCl3

Thanks

4 0

2 years ago

space capsules operate with an oxygen content of about 34%. assuming a total pressure of 780 mm Hg in the space capsule, what is

Lemur [1.5K]

265.2 mmHg is the partial pressure of oxygen in 780 mmHg of total pressure.

Explanation:

The partial pressure of a gas is defined as the individual pressure of the gas in total mixture. In an ideal gas all the constituent gases have partial pressure some of which will give total pressure of the gas.

The partial pressure of a gas is calculated by

total pressure x mole fraction of the gas.

Mole fraction of the oxygen present is 0.34 as it is 34% of the total gas.

$\frac{34}{100}$ = 0.34 is the mole fraction

Total pressure is given as 780 mm Hg

The partial pressure can be calculated using the above formula:

Putting the values in equation:

780 x 0.34

= 265.2 mm Hg is the partial pressure of oxygen.

7 0

2 years ago