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3 years ago

Which element, when combined with Iodine, would most likely form an ionic compound?

1 answer:

4 0

In an ionic compound the atoms are linked via ionic bonds. These are formed by the transfer of electrons from one atom to the other. The atom that loses electrons gains a positive charge whereas the atom that accepts electrons gains a negative. This happens in accordance with the octet rule wherein each atom is surrounded by 8 electrons

In the given example:

The valence electron configuration of Iodine (I) = 5s²5p⁵

It needs only one electron to complete its octet.

In the given options:

K = 4s¹

C = 2s²2p²

Cl = 3s²3p⁵

P = 3s²3p³

Thus K can donate its valence electron to Iodine. As a result K, will gain a stable noble gas configuration of argon while iodine would gain an octet. This would also balance the charges as K⁺I⁻ creating a neutral molecule.

Ans: Potassium (K)

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its the only logical answer, the others are either impossible or illogical

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3 years ago

When 38.1 grams of a certain metal at a temperature of 90.0°C is added to 100.0mL of water at a temperature of 17.6°C in a perfe

MakcuM [25]

Answer:

a. qm = 627.3 J

b. qw = 627.3 J

c. C₂ = 227.4 J/kg.°C

Explanation:

Since, the calorimeter is completely insulated. Therefore,

Heat Lost by Metal = Heat Gained by water

qm = qw

qm = m₁C₁ΔT₁

where,

qm = heat lost by metal = ?

m₁ = mass of water = (density)(volume) = (1000 kg/m³)(100 mL)(10⁻⁶ m³/1 mL)

m₁ = 0.1 kg

C₁ = specific heat capacity of water = 4182 J/kg.°C

ΔT₁ = Change in Temperature of Water = 19.1°C - 17.6°C = 1.5°C

Therefore,

qm = (0.1 kg)(4182 J/kg.°C)(1.5°C)

qm = 627.3 J

Since,

qm = qw

qw = 627.3 J

qm = m₂C₂ΔT₂

where,

m₂ = mass of metal = 38.1 g = 0.0381 kg

C₂ = specific heat capacity of metal = ?

ΔT₂ = Change in Temperature of metal = 90°C - 17.6°C = 72.4°C

Therefore,

627.3 J = (0.0381 kg)(C₂)(72.4°C)

(627.3 J)/(0.0381 kg)(72.4°C) = C₂

C₂ = 227.4 J/kg.°C

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3 years ago

Upon decomposition, a sample of sodium bicarbonate produces 0.0118 g of sodium carbonate, Na2CO3. How many grams of water does i

Naily [24]

Answer:

2.00 × 10⁻³ g

Explanation:

Step 1: Write the balanced decomposition reaction

2 NaHCO₃ ⇒ Na₂CO₃ + CO₂ + H₂O

Step 2: Calculate the moles corresponding to 0.0118 g of Na₂CO₃

The molar mass of Na₂CO₃ is 105.99 g/mol.

0.0118 g × 1 mol/105.99 g = 1.11 × 10⁻⁴ mol

Step 3: Calculate the moles of H₂O produced with 1.11 × 10⁻⁴ moles of Na₂CO₃

The molar ratio of Na₂CO₃ to H₂O is 1:1. The moles of H₂O produced are 1/1 × 1.11 × 10⁻⁴ mol = 1.11 × 10⁻⁴ mol.

Step 4: Calculate the mass corresponding to 1.11 × 10⁻⁴ moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

1.11 × 10⁻⁴ mol × 18.02 g/mol = 2.00 × 10⁻³ g

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3 years ago

How many moles of H2O are needed to react with a mole of 3.0 of Na?

AveGali [126]

Answer: 3.0 moles of water $(H_2O)$ are needed to react with a mole of 3.0 of Na

Explanation:

The balanced chemical reaction of sodium with water is as follows:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

According to stoichiometry:

2 moles of sodium $(Na)$ reacts with = 2 moles of water $(H_2O)$

Thus 3.0 moles of sodium $(Na)$ reacts with = $\frac{2}{2}\times 3.0=3.0$ moles of water $(H_2O)$

3.0 moles of water $(H_2O)$ are needed to react with a mole of 3.0 of Na

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3 years ago

When the same or different atoms combine together, you have a:

Alex73 [517]

Answer:

D.) molecule

<h3>When the same or different atoms combine together they for a molecule</h3>

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3 years ago