How many neutrons are probably in the nucleus of an element of atomic weight 197?

How many grams of NH3 can be produced from 12.0g of H2?

RSB [31]

Answer:

Balanced reaction:

3 H2 (g) + N2 (g) → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3. The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

8 0

2 years ago

Determine the charge on the unknown ion, X, in the compound Nax.

Leviafan [203]

Answer:

wax i think

Explanation:

i don't understand a thing you just wrote

3 0

2 years ago

If you need to extract a 50 ml aqueous solution with 50 ml of dichloromethane, what is the minimum size of a separatory funnel y

tatuchka [14]

The dichloromethane (DCM) has less density than water and also the polarity of water is much more than DCM. So the mixture of water and dichloromethane will always be a heterogeneous mixture. In the mixture dichloromethane will be always up of the water layer. The volume of the separatory funnel which contains the mixture of DCM and water must have to be more than the total volume of the liquids thus the volume of the funnel will be more than (50+50) = 100mL.

The caution have to consider during the separation are-

1. The separatory funnel have to shake well with lid and have to settle down for some times until the two liquid separated.

2. The lid should be open very slowly as the vapor pressure of DCM is more and it will float on the water.

3. After this the stopcock should be opened and slowly the water will come out first followed by DCM.

7 0

3 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

3 years ago

The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (

Mariulka [41]

Answer:

3.09kg

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol

Mass of C8H18 from the balanced equation = 2 x 114 = 228g

Converting 228g of C8H18 to kg, we obtained:

228/1000 = 0.228kg

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Converting 704g of CO2 to kg, we obtained:

704/1000 = 0.704kg

From the equation,

0.228kg of C8H18 produced 0.704kg of CO2.

Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2

6 0

3 years ago