‼️ASAP!!! BRAINLIEST!!‼️

If a chemical reaction produces 20.0 grams of product, but by stoichiometry it is supposed to have 25.0 grams of product; what i

Vitek1552 [10]

ThThe percentage yield is 80

4 0

2 years ago

What product is formed when benzene reacts with isobutyl chloride in the presence of AlCl3?

oee [108]

The product formed when benzene reacts with isobutyl chloride in the presence of AlCl3 is tertiary butyl benzene.

THE CHEMICAL REACTION AS

step 1 : CH₃ - CH - CH₂Cl + AlCl₃ → CH₃ - CH₂ -CH₂ - CH₂⁺ + AlCl⁻

step 2: CH₃ - CH₂ - CH₂ - CH₂⁺ ---H SHIFT--→ (CH₃)₃C⁺

PRODUCT

C₆H₆ + (CH₃)₃ C⁺ → C₆H₆ ( CH₃)₃ C⁺

Hence the product formed is tertiary butyl benzene.

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4 0

2 years ago

Gaseous compound Q contains only xenon and oxygen. When 0.100 gg of Q is placed in a 50.0 mLmL steel vessel at 0 ∘C∘C, the press

asambeis [7]

The question is incomplete, here is the complete question:

Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.229 atm. What is the likely formula of the compound?

A. XeO

B. $XeO_4$

C. $Xe_2O_2$

D. $Xe_2O_3$

E. $Xe_3O_2$

Answer: The chemical formula of the compound is $XeO_4$

Explanation:

To calculate the molecular mass of the compound, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{w}{M}RT$

where,

P = Pressure of the gas = 0.229 atm

V = Volume of the gas = 50.0 mL = 0.050 L (Conversion factor: 1 L = 1000 mL)

w = Weight of the gas = 0.100 g

M = Molar mass of gas = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = $0^oC=273K$

Putting value in above equation, we get:

$0.229\times 0.050=\frac{0.100}{M}\times 0.0821\times 273\\\\M=\frac{0.100\times 0.0821\times 273}{0.229\times 0.050}=195.4g/mol\approx 195g/mol$

The compound having mass as 195 g/mol is $XeO_4$

Hence, the chemical formula of the compound is $XeO_4$

5 0

3 years ago

When 15.0 mL of a 6.42×10-4 M sodium sulfide solution is combined with 25.0 mL of a 2.39×10-4 M manganese(II) acetate solution d

Artist 52 [7]

Answer:

Q = 3.59x10⁻⁸

Yes, precipitate is formed.

Explanation:

The reaction of Na₂S with Mn(CH₃COO)₂ is:

Na₂S(aq) + Mn(CH₃COO)₂(aq) ⇄ MnS(s) + 2 Na(CH₃COO)(aq).

The solubility product of the precipitate produced, MnS, is:

MnS(s) ⇄ Mn²⁺(aq) + S²⁻(aq)

And Ksp is:

Ksp = 1x10⁻¹¹= [Mn²⁺] [S²⁻]

Molar concentration of both ions is:

[Mn²⁺] = 0.015Lₓ (6.42x10⁻⁴mol / L) / (0.015 + 0.025)L = 2.41x10⁻⁴M

[S²⁻] = 0.025Lₓ (2.39x10⁻⁴mol / L) / (0.015 + 0.025)L = 1.49x10⁻⁴M

Reaction quotient under these concentrations is:

Q = [2.41x10⁻⁴M] [1.49x10⁻⁴M]

Q = 3.59x10⁻⁸

As Q > Ksp, the equilibrium will shift to the left producing MnS(s) the precipitate

8 0

4 years ago

Calculate the amount of heat needed to boil 120.g of acetic acid (HCH3CO2), beginning from a temperature of 16.7°C.

evablogger [386]

Answer:

The total amount of heat needed = 72.2116 kJ

Explanation:

Given that ;

the mass of acetic acid = 120.0 g

The initial temperature $T_1$ = 16.7 °C = (16.7 + 273.15 ) K = 298.85 K

The standard molar mass of acetic acid = 60.052 g/mol

Thus ; we can determine the number of moles of acetic acid;

number of moles of acetic acid = mass of acetic acid/ molar mass of acetic acid

number of moles of acetic acid = 120.0 g/ 60.052 g/mol

number of moles of acetic acid = 1.998 moles

For acetic acid:

The standard boiling point $T_2$ = 118.1 °C = ( 118.1 + 273.15 ) K = 391.25 K

The enthalpy of vaporization of acetic acid $\Delta H_{vap}$ = 23.7 kJ/mol

The heat capacity of acetic acid c = 2.043 J/g.K

The change in temperature Δ T = $T_2 - T_1$

Δ T = (391.25 - 289.85)K

Δ T = 101.4 K

The amount of heat needed to bring the liquid acetic acid at 16.7°C to its boiling point is ;

q = mcΔT

From our values above;

q = 120 g × 2.043 J/g.K × 101.4 K

q = 24859.2 J

q = 24859 /1000 kJ

q = 24.859 kJ

we have earlier calculated our number of moles o f acetic acid to be 1.998 moles;

Thus;

The needed amount of heat = $\Delta_{vap} *numbers \ of \ moles$

The needed amount of heat = $23.7 \ kJ/mol * 1.998 \ moles$

The needed amount of heat = 47.3526 kJ

Hence;

The total amount of heat needed = 24.859 kJ + 47.3526 kJ

The total amount of heat needed = 72.2116 kJ

4 0

3 years ago