1. I am in group two period three. What element am I?<br> A Ti<br> B, C<br> C. Ca<br> D. WE

Question: Draw a valid Lewis structure for the molecule CH3NO in which there are no nonzero formal charges on any of the atoms.

iVinArrow [24]

The mentioned molecule with formula, CH₃NO where no bond is found between N and O can be depicted as formamide is shown in the attachment below.

The formal charges = Number of valence electrons for the atom (V) - the number of electrons in lone pairs (N) - 1/2 (number of electrons in bond pairs, B)

FC = V - N - B/2

Thus, there is a need to calculate valence electrons, electrons in lone pairs, and the number of electrons in bond pairs for each atom in the mentioned molecule.

V or valence electrons on C = 4e, on H = 1e, on N = 5e, and on O = 6e.

N or electrons in lone pairs on C = 0e, on H = 0e, on N = 2e, and on O = 4e.

B or number of electrons in bond pairs for C = 8e, for H = 2e, for N = 6e, and for O = 4e.

Thus, the formal charges for each will be,

C = 4 - 0 - (8/2) = 0

H = 1 - 0 - (2/2) = 0

N = 5 - 2 - (6/2) = 0

O = 6 - 4 - (4/2) = 0

Lewis dot structure for the given molecule is given in the attachment below:

3 0

3 years ago

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Alex dragged a log across the yard in 30 the log weighted 400n she did 900 j of work how much power did I she have

In-s [12.5K]

The power used by Alex to drag the log across the yard is determined as 2,656 W.

The mass of the log is calculated as follows;

W = mg

m = W/g

m = (400)/9.8

m = 40.82 kg

<h3>Velocity of the log</h3>

K.E = ¹/₂mv²

v² = 2K.E/m

v² = (2 x 900)/(40.82)

v² = 44.096

v = 6.64 m/s

<h3>Power used by Alex</h3>

P = Fv

P = 400 x 6.64

P = 2,656 W

Learn more about power here: brainly.com/question/13881533

#SPJ1

5 0

1 year ago

A solution that is 0.20 m in hcho2 and 0.15 m in nacho2 find ph

Mashutka [201]

We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH

The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])

We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87

So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.

5 0

3 years ago

A football field is 120 yards by 53.333 yards. What is the area of the football field in acres if 1 acre=43560 ft?? Use correct

maw [93]

Answer:

1.3223 acres

Explanation:

a football field's area is 360 feet (120 yards) x 160 feet (53.333 yards) = 57,600 sq. feet

if the area of an acre is 43,560 sq. feet, then a football field in acres = 57,600 sq. feet / 43,560 sq. feet = 1.3223 acres

we can verify our answer by doing the same calculation in sq. yards:

football field = 120 yards x 53.33 yards = 6,400 sq. yards

an acre is 4,840 sq. yards

football field in acres = 6,400 sq. yards / 4,840 sq. yards = 1.3223 acres

8 0

3 years ago

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Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

<u>194 g/mol</u>

Explanation:

<u>1) Content of C:</u>

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

<u>2) Content of H</u>

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

<u>3) Content of N</u>

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

<u>4) Content of O</u>

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

<u>5) Ratios</u>

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

<u>6) Empirical formula:</u>

C₄H₅N₂O₁

<u>7) Calculate the approximate mass of the empirical formula:</u>

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago

1. I am in group two period three. What element am I? A Ti B, C C. Ca D. WE