Which invention has done the most to help predict weather and warn people of approaching storms?

Treatment of 1 mole of dimethyl sulfate with 2 moles of sodium acetylide results in the formation of propyne as the major produc

BartSMP [9]

Answer:

(a) appended underneath is the inorganic ion shaped in the reaction and the mechanism of its formation

(b) 2-butyne framed as a minor product is appeared in the connection. It is shaped when the monosodium subordinate of dimethylsulphoxide gets a hydrogen from the propyne and reacts again with monosodium methylsulphoxide.

(c) The major product framed when diethylsulphoxide is utilized, would be butyne and minor product would be 3-hexyne.

Explanation:

attached below is diagram

8 0

3 years ago

For the reaction 2 H 2 O ( g ) − ⇀ ↽ − 2 H 2 ( g ) + O 2 ( g ) the equilibrium concentrations were found to be [ H 2 O ] = 0.250

Bess [88]

<u>Answer:</u> The equilibrium constant for the given reaction is 4.224

<u>Explanation:</u>

We are given:

Equilibrium concentration of water = 0.250 M

Equilibrium concentration of hydrogen gas = 0.330 M

Equilibrium concentration of oxygen gas = 0.800 M

For the given chemical reaction:

$2H_2O(g)\rightleftharpoons H_2(g)+O_2(g)$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[H_2][O_2]}{[H_2O]^2}$

Putting values in above expression, we get:

$K_{eq}=\frac{0.330\times 0.800}{(0.250)^2}\\\\K_{eq}=4.224$

Hence, the equilibrium constant for the given reaction is 4.224

7 0

3 years ago

A solution of sodium hydroxide is standardized against potassium hydrogen phthalate. From the following data, calculate the mola

In-s [12.5K]

Answer:

0.7158 Mol/L

Explanation:

KHC8H4O4 + NaOH → H2O + KNaC8H4O4

KHC8H4O4 molar mass = 204.2212 g/mol

0.4536 g of KHC8H4O4 = 2.221 mMol (10^-3 mol)

volume = 31.26 mL - 0.23 mL = 31.03 mL

NaOH concentration = 2.221 mMol/31.03 mL = 2.221 Mol/31.03 L = 0.7158 Mol/L

6 0

4 years ago

2A1 (s) + 3C12 (g) --> 2AlCl3 (s) (balanced)

Dennis_Churaev [7]

Answer:

32.3%

Explanation:

Percent yield is defined as:

Actual yield (125.5g) / Theoretical Yield * 100

To find theoretical yield we have to find the moles of aluminium. As 2 moles of Al produce 2 moles of AlCl3, the moles of Al = Moles AlCl3.

With these moles we can find the mass assuming a 100% of yield (Theoretical Yield) as follows:

<em>Moles Al = Moles AlCl3 (Molar mass Al = 26.98g/mol)</em>

72g Al * (1mol / 26.98g) = 2.67 moles AlCl3

<em>Mass AlCl3 (Molar mass: 133.34g/mol)</em>

2.67 moles AlCl3 * (133.34g / mol) = 355.8g AlCl3

Percent Yield = 125.5g / 355.8g * 100 =

5 0

3 years ago

A galvanic cell consists of a iron electrode in 1 M Fe(NO3)2 and a copper electrode in 1 M Cu(NO3)2. What is the equilibrium con

Vladimir79 [104]

Answer:

Keq = 5.33*10²⁶

Explanation:

Based on the standard reduction potential table:

E°(Fe2+/Fe) = -0.45 V

E°(Cu2+/Cu) = +0.34 V

Since the reduction potential of copper is greater than iron, the former acts as the cathode and the latter as anode.

The half reactions are:

Cathode (Reduction): $Cu^{2+} + 2e^{-}\rightarrow Cu$

Anode (Oxidation): $Fe\rightarrow Fe^{2+}+ 2e^{-}$

------------------------------------------------------------------------------------------

Overall reaction: $Cu^{2+}+Fe\rightarrow Fe^{2+}+Cu$

The Gibbs free energy change at 25 C is related to the standard emf (E°) of the cell as well as the equilibrium constant K as:

$\Delta G^{0} = -RTlnK_{eq}=-nFE_{cell}^{0}$

here:

$E_{cell}^{0}= E_{cathode}^{0}-E_{anode}^{0}=0.34-(-0.45)=0.79V$

R = 8.314 J/mol-K

T = 25 C = 25+273 = 298 K

n = number of electrons involved = 2

F = 96500 Coulomb/mol e-

$8.314J/mol.K*298K*lnKeq= 2mole\ e^{-}*96500C/mole\ e^{-}*0.79V$

Keq = 5.33*10²⁶

6 0

4 years ago