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Phoenix [80]
3 years ago
15

There is no sodium found in nature in pure form due to its reactivity. What does this say about activation energy of the reactio

n of sodium and water compared to the combustion of wax in a candle?
Chemistry
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

The activation energy of a chemical reaction is the energy that is required to be supplied for a chemical reaction to take place. The activation energy for the reaction of sodium is low compared to the energy released such that the reaction of sodium and water is spontaneous resulting in the melting of the sodium into liquid form

The activation energy of a candle wax is much higher, requiring the ignition of the wick which burns and in turn melts the candle wax to release vapors that burns alongside the wick to produce sooth carbon dioxide, carbon monoxide and water vapor and release of heat energy which also fuels further combustion of the candle wax and wick

Therefore, the activation energy of the candle wax and wick which require the heat of direct flame from an ignited matches is higher than the activation energy of sodium placed in a medium of water that reacts spontaneously without heat application

Explanation:

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Which of these characteristics do these elements have in common?
Over [174]

C) They each form similar compounds when combined with oxygen.

Explanation:

The given elements have a common characteristics as they each form similar compounds when combined with oxygen.

Beryllium, magnesium and calcium all are located in the alkaline earth group.

Elements in the same group have the same chemical properties.

  • Density and melting point are physical properties of matter which varies down a group on the periodic table.
  • Their reaction with oxygen denotes a statement of their chemical properties.
  • Chemical properties of matter is a function of their valency.
  • Group two elements have a valency of two. They have two electrons in their outermost shell.
  • Therefore, they all form ionic bonds when they react with oxygen.

learn more:

Periodic table brainly.com/question/2690837

#learnwithBrainly

6 0
3 years ago
How many Liters of 0.50M HCl are needed to neutralize 0.050L of 0.101M Ba(OH)2?
Aleksandr-060686 [28]

Answer:

V_{HCl}=5.05x10^{-3}L

Explanation:

Hello,

In this case, since hydrochloric acid and barium hydroxide are in a 2:1 molar ratio, for the neutralization, the following moles equality must be obeyed:

2*n_{HCl}=n_{Ba(OH)_2}

In such a way, in terms of molarities and volumes, we can compute the required volume of hydrochloric acid as shown below:

2*M_{HCl}V_{HCl}=M_{Ba(OH)_2}V_{Ba(OH)_2}\\\\V_{HCl}=\frac{M_{Ba(OH)_2}V_{Ba(OH)_2}}{2M_{HCl}} =\frac{0.101M*0.050L}{2*0.50M} \\\\V_{HCl}=5.05x10^{-3}L

Besr regards.

8 0
3 years ago
Read 2 more answers
Identify whether each element is a halogen, a noble gas, or nonmetal only.
OLga [1]

Answer:

Astatine: Halogen

Nitrogen: Non-Metal

Krypton: Non-Metal, Noble Gas

Chlorine: Non-Metal

Sulfur: Non-metal

Explanation:

5 0
3 years ago
An ideal gas is made up of gas particles that
Hatshy [7]

Answer:

no it is mad out of gas

Explanation:

4 0
3 years ago
Read 2 more answers
Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
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