All categories

3 years ago

Susan and her friend found a piece of metal which they want to

Chemistry

1 answer:

joja [24]3 years ago

4 0

Answer:

The given metal is beryllium.

Explanation:

Given data:

Mass of metal = 74 g

Volume of metal = 40 cm³

Which metal is this = ?

Solution:

First of all we will calculate the density.

it is mass divided by volume.

d = m/v

d = 74 g/40 cm³

d = 1.85 g/cm³

In literature it is given that the density of beryllium is 1.85 g/cm³. Thus given metal is beryllium.

You might be interested in

What will be the end product for each electrode after electrolysis if the solution is concentrated aqueous sodium chloride ?

Eddi Din [679]

Answer:

Cathode: Hydrogen gas

Anode: Chlorine gas

Explanation:

① Write down the ions present in the electrolyte

Cations: Na⁺, H⁺

Anions: Cl⁻ , OH⁻

② Decide which ions are preferentially discharged.

These are the factors:

For the discharge of cations,

- Reactivity series

(The lower the position of the cation in the reactivity series, the easier it is to be discharged)

For the discharge of anions,

- Concentration effect (look at this first)

(The more concentrated the ion, the easier for it to be discharged)

- If solution is not concentrated (dilute), look at the position of the anion on the electrochemical series.

(The lower the position of the anion on the electrochemical series, the easier of it to be discharged)

In this case:

For cations, H⁺ ions are selectively discharged at the cathode as its position is lower than Na⁺ in the reactivity series.

Anion: Cl⁻ ions, being more concentrated, are selectively discharged at the anode.

☆For electrolysis,

Cation at the cathode (-ve terminal)

Anion at the anode (+ve terminal)

In summary, here's what happened at each electrode:

Cathode

- H⁺ selectively discharged

- ionic half equation: 2H⁺ (aq) +2e⁻ → H₂ (g)

Anode

- Cl⁻ ions selectively discharged

- ionic half equation: 2Cl⁻ (aq) → Cl₂ (g) + 2e⁻

3 0

3 years ago

4th one Mark the brainliest

ale4655 [162]

Answer:

1. The electronic configuration of X is: 1s2 2s2 sp6 3s2

2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

3. The formula of the compound form by X and Y is given as: XY

Explanation:

For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:

1s2 2s2 sp6 3s2

To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:

Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below

1s2 2s2 2p4

The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

The formula of the compound form by X and Y is given below :

X^2+ + Y^2- —> XY

Their valency will cancel out thus forming XY

4 0

4 years ago

Given that nitrogen forms three bonds with hydrogen to make <img src="https://tex.z-dn.net/?f=NH_%7B3%7D" id="TexFormula1" title

lbvjy [14]

Answer:

Three hydrogen atoms to form PH₃.

Explanation:

Hello!

In this case, since the elements belonging to the nitrogen family (N, P, As, Sb and Bi) show five valence electrons, because there are five electrons at their outer shell, it is clear that if phosphorous bonds with hydrogen, it is going to require the same amount of oxygen atoms (3) because elements having five valence electrons need 3 bonds in order to attain the octet (5+3=8).

Therefore the compound would be:

$PH_3$