Match the boxes helppppppp

a scuba diver's tank contains 0.29g of oxygen compressed into volume of 2.3L. What is the pressure in the tank at 9C?

viva [34]

Answer:

The answer to your question is P = 0.18 atm

Explanation:

Data

mass of O₂ = 0.29 g

Volume = 2.3 l

Pressure = ?

Temperature = 9°C

constant of ideal gases = 0.082 atm l/mol°K

Process

1.- Convert the mass of O₂ to moles

16 g of O₂ -------------------- 1 mol

0.29 g of O₂ ---------------- x

x = (0.29 x 1)/16

x = 0.29/16

x = 0.018 moles

2.- Convert the temperature to °K

Temperature = 9 + 273 = 282°K

3.- Use the ideal gas law ro find the answer

PV = nRT

-Solve for P

P = nRT/V

-Substitution

P = (0.018 x 0.082 x 282) / 2.3

-Simplification

P = 0.416/2.3

-Result

P = 0.18 atm

5 0

3 years ago

What is the probability of a<br> short plant in the cross<br> between a Tt and a tt plant?

kondaur [170]

75% would be your answer since the lower letter represents a shorter plant

7 0

2 years ago

During which month is the bird population in Georgia increasing the most

AlladinOne [14]

It always has birds, it's in the South.

3 0

2 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

2 years ago

Zinc metal is produced by heating solid zinc sulfide with solid wine witte resulting in liquid zinc and sulfur dioxide gas Write

elena-s [515]

Answer:

The balance chemical equation can be given as:

$ZnS(s)+ZnSO_4(aq)\overset{heat}\rightarrow 2Zn(s)+2SO_2(g)$

Explanation:

When zinc sulfide is allowed to react with zinc sulfate it gives zinc metal and sulfur dioxide gas as a product.The balance chemical equation can be given as:

$ZnS(s)+ZnSO_4(aq)\overset{heat}\rightarrow 2Zn(s)+2SO_2(g)$

According to reaction we can see when 1 mol of zinc sulfide reacts with 1 mol of zinc sulfate it forms 2 moles of zinc metal and 2 moles of sulfur dioxide gas.

4 0

2 years ago