My teacher hasn’t explained it well enough for me to actually understand it ;(

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

Oxygen gas, generated by the reaction 2 KClO 3 ---- 2 KCl + 3 O 2 , is collected over water at 27 C in a 2.00 L vessel at a tota

laila [671]

Answer:

The moles of KClO3 = 0.052 moles

Explanation:

Step 1: Calculate the pressure of oxygen gas

The oxygen has a total pressure (including water vapour) of 760 mmHg

The pressure of Oxygen = (760 - 26) mmHg

= 734 mmHg of water vapor

Step 2: Calculate the no of moles of oxygen

Using Ideal gas equation

P V = n R T

P = pressure of oxygen in N/m2 ( you should convert 734 mmHg to pascal or N/m2) = 97,858.6 N/m2 or pas

V = 2 litres = 0.002 m3

R = gas constant = 8.31

T= 27oC = 300 K

Applying this equation P V = n R T

97,858.6 x 0.002 = n x 8.31 x 300

n = 0.0785 mol of Oxygen

From the balanced equation

2 KClO 3 ---- 2 KCl + 3 O 2

3 moles of oxygen is produced from 2 moles KClO3

so 0.0785 mole of oxygen will be produced from x

x = (0.0785 x 2 ) / 3

x = 0.052 moles of KClO3

5 0

4 years ago

A 5.00g piece of metal is heated to 100.0°C, then placed in a beaker containing 20.0 of water at 10.0°C. The temperature of the

Ahat [919]

Answer:

This metal has a specific heat of 0.9845J/ g °C

Explanation:
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)

Step 2: Calculate specific heat
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)

- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =
-5 * (-85)(Cpmetal) = 418.4

Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C

This metal has a specific heat of 0.9845J/ g °C

7 0

3 years ago

If there are 9.6x1015 particles of sugar in a solution then how many moles of sugar are there?

julsineya [31]

Answer:1.6*10^-8

Explanation:

5 0

4 years ago

What is the molarity of a strong acid solution with a pH of 3?

Anna35 [415]

Answer:

Since HCl is a strong acid, it completely ionizes, and the pH of HCl in solution can be found from the concentration (molarity) of the H+ ions, by definition equal to 0.100 M. (The conjugate base of the acid, which is the chloride ion Cl–, would also have a concentration of 0.100 M.) The pH is thus –log(0.100) = 1.000.

Explanation:

6 0

3 years ago