All categories

3 years ago

Define covalent bond.

Chemistry

2 answers:

joja [24]3 years ago

3 0

Covalent bond/ a chemical bond that involves the sharing of electron pairs between atoms

telo118 [61]3 years ago

3 0

what................

You might be interested in

How many carbon atoms are represented in the formula for CaMg3(CO3)4

Galina-37 [17]

The formula is for 1 Calcium, 3 Magnesium, 4 Carbon, & 12 Oxygen. The 4 after the parenthesis of (CO3) distributes to each element.

Answer : There are 4 Carbon atoms in the formula.

3 0

2 years ago

Write word and balanced chemical equations to show the

Alex777 [14]

Answer:

chemical laboratory Pune Maharashtra India

5 0

3 years ago

A student fires a bow and arrow in gym class and all his arrows land close to each other, but not on the bullseye. This student

dimulka [17.4K]

A student fires a bow and arrow in gym class and all his arrows land close to eachother but not on the bullseye. this student could be said to be:

precise but not accurate

4 0

3 years ago

How many moles are there in 3.612 x 1024 formula units of Na2SO4?

vitfil [10]

Answer:

<h3>Theanswer is 6 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>6 moles</h3>

Hope this helps you

5 0

3 years ago

Sample A: 300 mL of 1M sodium chloride

yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

volume of sample B, V = 145 mL = 0.145 L

Molarity of sample B, C = 1.5 M

molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol

Molarity is given as;

$C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol$

The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g

The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g

The density of sample A $= \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L$

The density of sample B $= \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L$

Therefore, sample B contains the larger density

5 0

2 years ago