If an oxygen has 1s2 2s2 2p4 , how many more electrons should it have to be octet .

Whichof the following is another name for science

snow_lady [41]

Field or discipline
hope it helped

8 0

3 years ago

Which of the following is most likely to lose electrons in an ionic compound?

ANTONII [103]

Answer:

a. Lead (Pb)

Am 100% sure

Hope it helps

please mark me as brainliest

4 0

3 years ago

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

7 0

3 years ago

The specific heat of a certain type of metal is 0.128J/(g.°C). What is the final temperature if 305 J of hear is added to 28.8 g

klasskru [66]

Answer: Temperature final = 103 °C

Explanation: To solve for final temperature we use the equation of heat:

Q= mc∆T

Next derive the equation to find final temperature

Q = mc(T final - T initial)

Q / mc = T final - T initial

Transpose T initial and change the sign so that T final will be left.

T final = Q / mc + T initial

Substitute the values:

T final = 305 J / 28.8 g x 0.128 J/(g°C)

= 305 J / 3.6864 J/°C

= 82.7 + 20.0°C

= 103 °C final temperature.

8 0

3 years ago

Nicotine, a component of tobacco, is composed of c, h, and n. a 4.200-mg sample of nicotine was combusted, producing 11.394 mg o

Rudiy27

Answer:

Empirical Formula = C₅H₇N₁

Solution:

Data Given:

Mass of Nicotine = 4.20 mg = 0.0042 g

Mass of CO₂ = 11.394 mg = 0.011394 g

Mass of H₂O = 3.266 mg = 0.003266 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.011394 ÷ 0.0042) × (12 ÷ 44) × 100

%C = (2.7128) × (12 ÷ 44) × 100

%C = 2.7128 × 0.2727 × 100

%C = 73.979 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.003266 ÷ 0.0042) × (2.02 ÷ 18.02) × 100

%H = (0.7776) × (2.02 ÷ 18.02) × 100

%H = 0.7776 × 0.1120 × 100

%H = 8.709 %

%N = 100% - (%C + %H)

%N = 100% - (73.979 % + 8.709%)

%N = 100% - 82.688%

%N = 17.312 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 73.979 ÷ 12.01

Moles of C = 6.1597 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 8.709 ÷ 1.01

Moles of H = 8.6227 mol

Moles of N = %N ÷ At.Mass of O

Moles of N = 17.312 ÷ 14.01

Moles of N = 1.2356 mol

Step 3: Find out mole ratio and simplify it;

C H N

6.1597 8.6227 1.2356

6.1597/1.2356 8.6227/1.2356 1.2356/1.2356

4.985 6.978 1

≈ 5 ≈ 7 1

Result:

Empirical Formula = C₅H₇N₁

6 0

3 years ago