All categories

1 year ago

State and explain the trend in electronegativity values across period 3 from sodium to chlorine

1 answer:

3 0

The trend in electronegativity values across period 3 from sodium to chlorine is that sodium with 11 protons is the least electromagnetic element and chlorine with 17 protons is the most electronegative element.

<h3>What is electronegativity?</h3>

It should be noted that electronegativity simply means a chemical property that describes the tendency of an atom to attract elements towards it.

In this case, the trend in electronegativity values across period 3 from sodium to chlorine is that sodium with 11 protons us the least electromagnetic element and chlorine with 17 protons is the most electronegative element.

Learn more about electronegativity on:

brainly.com/question/24977425

#SPJ12

You might be interested in

A hypothetical wave has 4.6 J of energy. What is its hypothetical, approximate frequency?

Mariulka [41]

Answer:

6.94 × 10^33Hz

Explanation:

E = hf

Where;

E = Energy of wave (J)

h = Planck's constant (6.626 × 10^-34J/s)

f = frequency (Hz)

According to the information provided in this question, the hypothetical energy of the wave is 4.6J

Hence, using E = hf

4.6 = 6.626 × 10^-34 × f

f = 4.6 / 6.626 × 10^-34

f = 0.694 × 10^34

f = 6.94 × 10^33Hz

3 0

3 years ago

If the point of the nail can be approximated as a circle with a radius 2.00×10^-3m What is the pressure in MPa exerted on the wa

Natasha_Volkova [10]

Answer:

8.28 MPa

Explanation:

From the question given above, the following data were obtained:

Radius (r) = 2×10¯³ m

Force applied (F) = 104 N

Pressure (P) =?

Next, we shall determine the area of the nail (i.e circle). This can be obtained as follow:

Radius (r) = 2×10¯³ m

Area (A) of circle =?

Pi (π) = 3.14

A = πr²

A = 3.14 × (2×10¯³)²

A = 3.14 × 4×10¯⁶

A = 1.256×10¯⁵ m²

Next, we shall determine the pressure. This can be obtained as follow:

Force applied (F) = 104 N

Area (A) = 1.256×10¯⁵ m²

Pressure (P) =?

P = F / A

P = 104 / 1.256×10¯⁵

P = 8280254.78 Nm¯²

Finally, we shall convert 8280254.78 Nm¯² to MPa. This can be obtained as follow:

1 Nm¯² = 1×10¯⁶ MPa

Therefore,

8280254.78 Nm¯² = 8280254.78 Nm¯² × 1×10¯⁶ MPa / 1 Nm¯²

8280254.78 Nm¯² = 8.28 MPa

Thus, the pressure exerted on the wall is 8.28 MPa

3 0

2 years ago

Molecular compounds are capable of forming multiple covalent bonds. the compound below that contains a triple bond is:

alexira [117]

The answer is CO(g).

4 0

1 year ago

Chemitey half equations

shepuryov [24]

Isn't it a because in b at the start of the equation the E in Fe just disappeared

8 0

2 years ago

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago