How many moles of xenon gas are necessary to exert a 0.64atm pressure in an 8.5L container at 300.0K?

Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a

Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

Pressure = 1.50 *10^6 Pa

The pressure reduce to 0.950 *10^6 Pa

The temperature stays constant at 200 °C

Step 2: Calculate the volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.50 * 10^6 Pa

⇒with V1 = the initial volume = 25 L

⇒with P2 = the final pressure = 0.950 * 10^6 Pa

⇒with V2 = the final volume = TO BE DETERMINED

1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2

V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)

V2 = 39.5 L = 0.0395 m³

The final volume is 39.5 L = 0.0395 m³

3 0

2 years ago

If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le

ElenaW [278]

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

$m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g$

Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

$m=\frac{92.5}{100}\times 0.5940=0.54945 g$

1 g of lead obtain from $\frac{1}{0.54945}=1.82$ grams of ore.

Thus, 2041.2 g of lead obtain from:

$2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg$

Therefore, mass of ore required to make lead sphere is 3.715 kg.

6 0

2 years ago

Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.

WARRIOR [948]

Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer

5 0

2 years ago

. Divide 94.20 g by 3.167 22 mL.

jeka94

94.20 g/3.16722 mL = 29.74 g/mL

The ratio of mass to volume is equal to the substance's density. Thus, 29.74 g/mL is the density of whatever substance it may be. Density does not change for incompressible matter like solid and some liquids. Although, it may be temperature dependent.

7 0

2 years ago

Read 2 more answers

Match the following names of glassware with what you would use them for.(1) Glassware used to accurately transfer small volumes.

Andrei [34K]

Answer:

(1)=(A), (2)=(B), (3)=D, (4)=C, (5)=E, (6)=F

Explanation:

(1) Glassware used to accurately transfer small volumes = (A) Graduated pipette, that is basically a glass tube with graduation of different volumes to be dispensed.

(2) Glassware used to accurately transfer a small, single volume = (B) Volumetric pipette, that is a glass tube with a central glass bulb and is used to dispense accurately an unique volume of liquid everytime.

(3) Glassware to deliver a volume not known in advance = (D) Buret (or burette), that is used to dispense slowly a volume of liquid when a titration process is needed

(4) Glassware best used when greater access to the contents is needed = (C) Beaker, that is basically a very open glass cylinder with a spout

(5) Glassware used to prevent splashing or evaporation = (E) Erlenmeyer flask, that has a small open at the top and is useful when the liquid needs to be swirled as, for example, during a titration.

(6) Glassware used to make accurate solutions = (F) Volumetric flask, that has a long slim neck that provides a higher accuracy when a exact volume of liquid needs to be used for preparation of a solution.

8 0

2 years ago