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Elena-2011 [213]
3 years ago
15

PLEASE HELP!

Chemistry
1 answer:
svetoff [14.1K]3 years ago
8 0

Answer:

period 3 and group 3

Explanation:

I'm saying group 3 because that is how I learnt it at school, but if you count it then it's in group 13.

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Which element would you expect to be more metallic?<br> (a) Ca or Rb (b) Mg or Ra (c) Br or I
svlad2 [7]

Explanation:

When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.

As calcium (Ca) is a group 2A element and rubidium (Rb) is a group 1A element. Hence, Rb being an alkali metal is more metallic in nature than calcium (alkaline earth metal).

Both magnesium (Mg) and radium (Ra) are group 2A elements. And, when we move down a group then as the size of element increases so, it becomes easy of the metal atom to lose an electron.

As a result, there occurs an increase in metallic character of the element. Hence, Radium (Ra) is more metallic in nature than magnesium (Mg).

Also, both bromine and iodine are group 17 elements. Since, both of them are non-metals and non-metallic character increases on moving down the group.

Therefore, bromine (Br) is more metallic than iodine.

7 0
3 years ago
Which metal reacts quickly with cold water only when it is finely powdered?
scoundrel [369]
Most likely it is magnesium
7 0
3 years ago
Read 2 more answers
A 30ml sample of a liquid has a mass of 50 grams. What is the density of the liquid?
mash [69]

Answer:

<h2>Density = 1.67 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

From the question

mass = 50 g

volume = 30 mL

Substitute the values into the above formula and solve for the density

That's

Density =  \frac{50}{30}  \\  =  \frac{5}{3}  \\  = 1.66666...

Wr have the final answer as

<h3>Density = 1.67 g/mL</h3>

Hope this helps you

5 0
3 years ago
How many grams of bromine are required to react completely with 37.4 grams aluminum chloride?
Gekata [30.6K]

Answer:

None  

Explanation:

Cl₂ is above Br₂ in the activity series.

Bromine will not displace chlorine from its salts.

The reaction will not occur.

8 0
3 years ago
Read 2 more answers
The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
3 years ago
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