All categories

3 years ago

Which of the following does not form a diatomic molecule?

Chemistry

2 answers:

Vinvika [58]3 years ago

8 0

d. copper is the correct answer to the following question.

Murrr4er [49]3 years ago

7 0

The correct answer among the choices given is option D. Copper does not form a diatomic molecule. A diatomic molecule has only two atoms. Diatomic molecules are molecule that are composed of two same atoms which are bonded covalently. An example are oxygen, it exist as O2 and chlorine, it exist as Cl2.

You might be interested in

A solution at 25 degrees Celslus 1.0*10^ -5 MH 3 O^ + . What the concentration of OH^ - in this solution?

dexar [7]

Answer:

Explanation:

-log(1.0x10^-5) = pH

pH + pOH = 14 (rearrange it)

OH- = 10^-pOH = 1.0 x 10^-9

- Hope that helped! Let me know if you need further explantion.

7 0

3 years ago

How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP

Rom4ik [11]

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

Solution :

At STP,

22.4 L volume of ethane present in 1 mole of ethane gas

64.28 L volume of ethane present in $\frac{64.28L}{22.4L}\times 1mole=2.869moles$ of ethane gas

And, as we know that

1 mole of ethane molecule contains $6.022\times 10^{23}$ molecules of ethane

2.869 moles of ethane molecule contains $2.869\times 6.022\times 10^{23}=17.277\times 10^{23}$ molecules of ethane

Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

4 0

3 years ago

Read 2 more answers

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in

In-s [12.5K]

Answer:

3.6667

Explanation:

For helium gas:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${5.6}\times {3.0}={P_2}\times {9.0} atm$

${P_2}=\frac {{5.6}\times {3.0}}{9.0} atm$

${P_1}=1.8667\ atm$

The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm

For Neon gas:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${3.6}\times {4.5}={P_2}\times {9.0} atm$

${P_2}=\frac {{3.6}\times {4.5}}{9.0} atm$

${P_1}=1.8\ atm$

The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm

Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.

6 0

3 years ago

An aqueous solution is 40.0 % by mass hydrochloric acid, HCl, and has a density of 1.20 g/mL. The mole fraction of hydrochloric

vivado [14]

Answer:

The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

Molar concentration = mass of HCl/ molar mass of HCl

Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.

We can extract the mass of the solution from its density which is 1.2g/mL

We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

The mass of HCl Present is 40% of 1.2 = 0.48 g.

The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol

Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3

7 0

3 years ago

If your front lawn is 18.0 feet wide and 20.0 feet long. And each square foot of lawn accumulates 1450 new snow flakes every min

creativ13 [48]

To solve this problem, we begin by first calculating the area of the front lawn. The length and width of the lawn was given and the area of a rectangle is given by the formula: Area = length x width. Thus, the area of the front lawn can be obtained by multiplying 18 ft by 20 ft, wherein we get 360 ft^2 as the area.

Second, the problem indicated that each square foot of lawn accumulates 1450 new snow flakes per minute. This can be translated into the expression 1450 snow flakes/ (minute·ft^2). In this way, we can convert it to units of mass (kg). Afterwards, we simply need to multiply it to the area of the lawn and convert minute to hour. The following expression is then used:

1450 snow flakes/ (minute·ft^2) x 1.90 mg/snow flake x 1 g/1000 mg x 1kg/1000 g x 360 ft^2 x 60 minutes/hour = 59.508 kg snow flake/hour

It is then calculated that 59.508 kg of snow flake accumulates in the lawn every hour.

6 0

3 years ago