Which of these is an empirical formula?

The poison dart from the jungle can kill its prey with a slim from its skin. this is an example of a

nikdorinn [45]

This is an example of a chemical property.

8 0

4 years ago

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

How many things do controlled experiments attempt to test at one time? 0 1 3 2

Hunter-Best [27]

Answer:

1

Explanation:

your can only test one thing at a time

5 0

3 years ago

Read 2 more answers

Whta gasses are in Neptune?

Natasha_Volkova [10]

Neptune's atmosphere is composed up of many gases. These gases include Hydrogen (80%), Helium (19%), and Methane (1.5%). Some of the minor gases include trace amounts of Hydrogen Deuteride and Ethane.

6 0

3 years ago

The vapor pressure of CCl3F at 300 K is 856 torr. If 11.5 g of CCl F is enclosed in a 1.0-L container, will any liquid be pres-

Lemur [1.5K]

Answer:

Yes, there will be liquid present and the mass is 5.19 g

Explanation:

In order to do this, we need to use the equation of an ideal gas which is:

PV = nRT (1)

Where:

P: Pressure

V: Volume

n: number of moles

R: gas constant

T: Temperature

we know that the pressure is 856 Torr at 300 K. So, if we want to know if there'll be any liquid present, we need to calculate the moles and mass of the CCl3F at this pressure and temperature, and then, compare it to the initial mass of 11.5 g.

From (1), solving for moles we have:

n = PV/RT (2)

Solving for n:

P = 856/760 = 1.13 atm

R = 0.082 L atm / mol K

n = 1.13 * 1 / 0.082 * 300

n = 0.0459 moles

Now, the mass is:

m = n * MM (3)

The molar mass of CCl3F reported is 137.37 g/mol so:

m = 0.0459 * 137.37

m = 6.31 g

Finally, this means that if we put 11.5 g of CCl3F in a container, only 6.31 g will become gaseous, so, this means it will be liquid present, and the mass is:

m = 11.5 - 6.31

m = 5.19 g

7 0

4 years ago