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Citrus2011 [14]
3 years ago
12

How do the bonds differ in each type?

Chemistry
2 answers:
VashaNatasha [74]3 years ago
7 0

Answer:

Bonds basically differs with each other due to sharing of electrons .

Explanation:

There are majorly three kinds of bonds

1. Ionic bonds which forms due to an element donate an electron to another element completely .

2. covalent bonds which forms with the mutual sharing of electrons b/w two atoms .

3. metallic bonds which forms b/w metals & they share electrons due to electron negativity difference b/w two atoms or elements

hichkok12 [17]3 years ago
7 0
The two main types of chemical bonds are ionic and covalent bonds. An ionic bond essentially donates an electron to the other atom participating in the bond, while electrons in a covalent bond are shared equally between the atoms. ... Covalent bonds form between two nonmetals
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An element's atomic number is 112. How many electrons would an atom of this element have?
PIT_PIT [208]

Without any ionization, the element (Cn) would have 112 electrons because the atomic number of an element is the number of protons the element has and a neutral element has the same number of electrons as it does protons.

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describe how to put a food you enjoy such as hamburger pizza or muffin identify one physical change that takes place as this foo
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How many milliliters of 0.100 m naoh are required to neutralize 9.00 ml of 0.0500 m hcl?
BabaBlast [244]
V  ( NaOH ) = mL ?

M ( NaOH ) = 0.100 M

V ( HCl ) = 9.00 mL / 1000 => 0.009 L

M ( HCl ) = 0.0500 M

number of moles HCl:

n = M x V

n = 0.009 x 0.0500 => 0.00045 moles HCl

mole ratio:

<span>HCl + NaOH = NaCl + H2O
</span>
 1 mole HCl ---------------- 1 mole NaOH
 0.00045 moles HCl ----- ??

0.00045 x 1 / 1 => 0.00045 moles of NaOH

M = n / V

0.100 = 0.00045 / V

V = 0.00045 / 0.100

V = 0.0045 L

1 L ------------ 1000 mL
0.0045 L ----- ??

0.0045 x 1000 / 1 => 4.5 mL of NaOH



6 0
4 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
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