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3 years ago

What is the most likely type of bond formed between element x and iodine?

1 answer:

5 0

The most likely bond between element X and Iodine would be an ionic, or electrovalent, bond. Iodine has seven electrons in its outer shell, also known as the valence shell. To become perfectly stable, it needs only a single electron from another element. Hence no sharing of electron takes place (usually), which is the condition required for it to be covalent bonding. Hence it's most likely an ionic bonding/

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Two charged particles repel each other with a force of F.

mihalych1998 [28]

Answer:

it would increase to twice what it was

3 0

2 years ago

What was The US Army trying to slove about lake Lenore in 1947

Leviafan [203]

Answer:

1947 the United States Army had an excess of metallic sodium left over from World War II and determined that the alkaline waters of Lake Lenore would be a good spot to dump and neutralize the acidic element, which reacts with water with intense explosions.

Explanation:

there ya go

5 0

2 years ago

What is the name of(NH4)3PO4

Tju [1.3M]

Answer:

What is the name of the compound (NH4) 3PO4? (NH)4+ is an ammonium radical and (PO4) 3- is a phospate radical. When we start writing the compound,the valency of phosphate goes to ammonium as subscript,and that of ammonium goes to phosphate. Hence the formula (NH4)3 PO4 and it's name is ammonium phospate.

Explanation:

5 0

3 years ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

2 years ago

What is the electron configuration of argon

PilotLPTM [1.2K]

Atomic number - 18
1s2,2s2,sp6,3s2,3p6
noble configuration is [Ne]3s2,3p6

7 0

3 years ago