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4 years ago

The ion that has the same e- arrangement as Ar:

Chemistry

2 answers:

alina1380 [7]4 years ago

7 0

Well, I recommend reading the text :P

Alisiya [41]4 years ago

5 0

the answer is E- all of these

to check if these ions have the same e- configuration, check them in the periodic table and see where they land when you take or add electrons.

for example, K is one space away to be like Ar, which is just one space behind. having a positive charge +1, means that it has one less electron. if you remove this electron, it is like Ar because is one space.

another example-> S-2 means you have add two electrons because charge is negative, so you add. two spaces after S land in Ar. so it is same e- configuration as Ar.

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Arrange the following molecules in order of increasing bond polarity (highest bond polarity at the bottom):

Tom [10]

NF3– 0.94– third
NCl3–0.12– second
NBr3–0.08– first
CF4–1.43– fourth

NBr3—NCl3—NF3—CF4
Lowest. Highest

4 0

3 years ago

What is the appearance of H2o?

telo118 [61]

Answer:

Water

Explanation:

one molecule of water has two hydrogen atoms covalently bonded to a single oxygen atom. Water is a tasteless, odorless liquid at ambient temperature and pressure. Which is known to be crystal blue.

6 0

3 years ago

Read 2 more answers

A chemistry student is given 600. mL of a clear aqueous solution at 37.° C. He is told an unknown amount of a certain compound X

barxatty [35]

Answer:

Yes, it is 14. g of compound X in 100 ml of solution.

Explanation:

The relevant fact here is:

the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

84. g solute / 600 ml solution = y / 100 ml solution

⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

The answer is 14. g of solute per 100 ml of solution.

7 0

3 years ago

A 0.04328 g sample of gas occupies 10.0-mL at 294.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C

stepan [7]

Answer:

The molecular formula of the gas sample = $C_2Cl_2$

Lewis structure is shown in the image below.

The geometry around each carbon atom is linear.

Explanation:

Given that:

Temperature = 294.0 K

V = 10.0 mL = 0.01 L ( 1 mL = 0.001 L )

Pressure = 1.10 atm

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

1.10 atm × 0.01 L = n ×0.0821 L atm/ K mol × 294.0 K

⇒n = 0.0004557 mol

Given, mass = 0.04328 g

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0004557\ mole= \frac{0.04328\ g}{Molar\ mass}$

Molar mass of the gas sample = 94.9747 g/mol

Given that:-

% of C = 25.305

Molar mass of C = 12.0107 g/mol

% moles of C = 25.305 / 12.0107 = 2.1069

% of Cl = 74.695

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 74.695 / 35.453 = 2.1069

Taking the simplest ratio for C and Cl as:

2.1069 : 2.1069 = 1 : 1

The empirical formula is = $CCl$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 35.5 = 47.5 g/mol

Molar mass = 94.9747 g/mol

So,

Molecular mass = n × Empirical mass

94.9747 = n × 47.5

⇒ n = 2

The molecular formula of the gas sample = $C_2Cl_2$

Also,

Valence electrons of carbon = 4

Valence electrons of Chlorine = 7

The total number of the valence electrons = 4*2 + 7*2 = 22

The Lewis structure is drawn in such a way that the octet of each atom in the molecule is complete. So,

The Lewis structure is shown in the image below.

According to the theory, the atoms will form a geometry in such a way that there is minimum repulsion and maximum stability. The carbon atoms are sp hybridized.

So, it is of linear shape.

3 0

4 years ago

How many calories is 130. joules?

skad [1K]

1 Cal ---------- 4.184 J
? Cal ---------- 130.0 J

130.0 x 1 / 4.184 => 31.07 Cal

hope this helps!

6 0

4 years ago