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aleksklad [387]
3 years ago
13

Described the role of electrons in the formation of a covalent bond​

Chemistry
1 answer:
My name is Ann [436]3 years ago
7 0

Sharing of two electrons make a <u>Covalent </u>bond.

<u>Explanation: </u>

Attractions among the atoms bring them together. So the electrons from each of the atoms are attracted towards the nucleus of those two atoms, that “share” the electrons produces a covalent bond.

It is also named as molecular bond,  a bond that entails the sharing of a pair of electrons among the atoms. When the atoms share the electrons among themselves, it produces a molecule, which is more stable than the atom.

If the attractions between the atoms are strong enough and if every atom has enough space for the electrons in its outermost energy level then there occurs covalent bonding. So electrons are very important in the covalent bond formation.

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A system receives 575 ) of heat and delivers 425 ) of work. Calculate the change in the internal energy. AE, of the system.​
vfiekz [6]

Answer:

ΔE = 150 J

Explanation:

From first law of thermodynamics, we know that;

ΔE = q + w

Where;

ΔE is change in internal energy

q is total amount of heat energy going in or coming out

w is total amount of work expended or received

From the question, the system receives 575 J of heat. Thus, q = +575 J

Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.

Thus;

ΔE = 575 + (-425)

ΔE = 575 - 425

ΔE = 150 J

7 0
3 years ago
Chemically combined substances, such as CO2, NH3, and H20 are called.....
Alex787 [66]
It is d. compounds because they have more than one element.
8 0
3 years ago
Read 2 more answers
Find the weight of HNO_3 present in 20ml, 0.30 N
yanalaym [24]

Answer:

mass of HNO₃ = 0.378 g

Explanation:

Normality = Molarity * number of equivalents

Molarity = Normality/number of equivalents

normality of HNO₃ = 0.30 N, Volume = 20 mL

HNO₃ ionizes in the following way:

HNO₃(aq) ----> H⁺ + NO₃⁻

Therefore, number of equivalents for HNO₃ is 1

molarity of HNO₃ = 0.30/1 =0.30 mol/dm³

Using the formula, molarity = number of moles/volume in liters

number of moles = molarity * volume

Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL

number of moles = 0.006 moles

From the formula, mass = number of moles * molar mass

molar mass of HNO₃ = 63.0 g/mol

mass = 0.006 * 63

mass of HNO₃ = 0.378 g

6 0
3 years ago
If water is added to a 0.70 molar solution of CuSO4 what will change
lana [24]

Answer:

B. Molarity will decrease

Explanation:

Molarity is one of the measures of the molar concentration of a solution. It is calculated by dividing the number of moles of the solute by the volume of the solvent. This means that the higher the amount of solute in relation to the volume of solvent, the higher the molarity of that solution.

In essence, adding water to a solution dilutes it i.e it increases the solvent's volume in relation to the solute, causing the molarity to decrease. In a nutshell, diluting a solution (by adding water or more solvent) causes the molarity of such solution to decrease. For example, if water is added to a 0.70 molar solution of CuSO4, the molarity of the solution will DECREASE.

7 0
3 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
3 years ago
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