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3 years ago

Which of the following compounds most likely has the least bond energy?

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

4 0

Answer:

I2; I–I bond length = 266 pm

Explanation:

Bond length is inversely related to bond strength. The longer the bond length, the weaker the bond. The shorter the bond length the stronger the bond. A large bond distance implies that there is poor interaction between the atoms involved in the bond. A long bond distance or bond length may even indicate the absence of covalent interaction between the atoms involved.

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Calcula la densidad de un cuerpo que tiene la masa de 12g y un volumen de 3cm^3 dar el resultado en unidades del SI

kvv77 [185]

The density is 4 g/cm³ or 4000 kg/m³.

Density = mass/volume = 12 g/3 cm³ = 4 g/cm³

The measurement of 4 g/cm³ is already in <em>SI units</em>.

In SI <em>bas</em>e units,

Density = (4 g/1 cm³) × (1 kg/1000 g) × (100 cm/1 m)³ = 4000 kg/m³

3 0

3 years ago

What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,

boyakko [2]

The rate constant of a reaction : 8.3 x 10⁻⁴

<h3>Further explanation</h3>

Given

rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1

Required

the rate constant

Solution

For aA + bB ⇒ C + D

Reaction rate can be formulated:

$\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}$

the rate constant : k =

$\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}$

8 0

2 years ago

Read 2 more answers

What is the solubility of ethylene (in units of grams per liter) in water at 25 °C, when the C2H4 gas over the solution has a pa

vredina [299]

<u>Answer:</u> The solubility of ethylene gas in water is $9.16\times 10^{-2}g/L$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{C_2H_4}=K_H\times p_{C_2H_4}$

where,

$K_H$ = Henry's constant = $4.78\times 10^{-3}mol/L.atm$

$C_{C_2H_4}$ = molar solubility of ethylene gas = ?

$p_{C_2H_4}$ = partial pressure of ethylene gas = 0.684 atm

Putting values in above equation, we get:

$C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L$

Converting this into grams per liter, by multiplying with the molar mass of ethylene:

Molar mass of ethylene gas = 28 g/mol

So, $C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L$

Hence, the solubility of ethylene gas in water is $9.16\times 10^{-2}g/L$

6 0

3 years ago

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO 3 ( s ) + 2

denis23 [38]

Answer:

Mass of formed CaCl₂ is 14.9 g

Mass of CaCO₃ that remains after the reaction is 14.5 g

Explanation:

We determine the reaction:

CaCO₃ (s) + 2HCl (aq) ⟶ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

We have the mass of both reactants, so we have to find out which is the limiting; firstly we convert the mass to moles

28 g . 1mol / 100g = 0.28 moles of CaCO₃

10 g . 1mol /36.45g =0.27 moles of HCl

Ratio is 1:2. Let's make a rule of three:

1 mol of carbonate needs 2 moles of acid, to react

Then, 0.28 moles must react with the double of moles, of acid.

Therefore the HCl is the limiting reactant, and the reagent in excess is the salt.

Ratio is 2:1. The rule of three will be:

2 moles of acid needs 1 mol of salt to react

Then, 0.27 moles of acid, will react with (0.27 .1 ) / 2 = 0.135 moles of salt

We have 0.28 moles of carbonate so (0.28 - 0.135) = 0.145 moles of salt will remain after the reaction is complete.

We convert the moles to mass → 0.145 mol . 100 g/1mol = 14.5 g

We said, that the limiting reagent was the HCl so we can work with the products, now. Ratio is 2:1

2 moles of HCl can produce 1 mol of CaCl₂

Then, 0.27 moles of HCl will produce (0.27 . 1)/ 2 = 0.135 moles of chloride

We convert the moles to mass → 0.135 mol . 110.98 g/1mol = 14.9 g

8 0

3 years ago

Is pepperoni pizza an element ?

trasher [3.6K]

No because pepperoni pizza is an object.

6 0

3 years ago

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