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Musya8 [376]
3 years ago
10

What does an aluminum atom become when it loses its three valence electrons? a?

Chemistry
1 answer:
Artist 52 [7]3 years ago
6 0
Aluminum atomic number 13 becomes Neon with an atomic number of 10 when it loses three valence electrons.
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Select the correct answer.
sladkih [1.3K]
C not sure 100 percent but my best guess
7 0
3 years ago
Scientists believe that interactions between atoms depend mostly on the arrangement of the outermost electrons in each atom. Thi
vova2212 [387]
I’m pretty sure it’s in a group (column) of the period table. Hope this helps :)))
5 0
2 years ago
Predict the aldol product when the following ketone undergoes self-condensation in the presence of NaOH. Do not show the dehydra
Anit [1.1K]

Answer:

β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal.

Explanation:

When acetaldehyde is treated with dil.NaOH it undergoes self condensation as it contains alpha-hydrogen atom in its compound forming β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal. This compound upon further heating will eliminate a molecule of water forming aldol condensation product namely Crotonaldehyde Or But-2-en-al. see the diagram attached.

5 0
3 years ago
Combustion of 6.90g of this compound produced 13.8g of CO2 and 5.64g of H2O. What is the empirical formula of the unknown compou
Alla [95]
The empirical formula for the unknown compound would be: C2H4O (2 molecules of Carbon, 4 molecules of Hydrogen, and 1 molecule of Oxygen)
4 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
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