Minerals are grouped according to their

A 1.3 g sample of a substance is heated from 0°C to 45°C and is found to have absorbed 45 j of heat. What is the specific heat o

wlad13 [49]

Q=mcat

45=(1.3)(c)(45)

45=58.5c

.769=c

~.77

The answer is .77

5 0

3 years ago

An orgo lab student named Bob wanted to synthesize an isoamyl acetate ester because it smells like bananas and he really likes t

mihalych1998 [28]

Answer:

The answer is "11.07 g".

Explanation:

Isoamyl alcohol is a reagent restriction

Isoamyl alcohol Moles:

$= \frac{7.5}{88.15} \\\\ =0.085$

Moles only with the shape of isoamyl acetate are equivalent to numbers.

Isoamyl acetate grams:

$= 0.085 \times 130.19\\\\ = 11.07 \ g$

3 0

3 years ago

n a gas-phase equilibrium mixture of SbCl5, SbCl3, and Cl2 at 500 K, pSbCl5 = 0.17 bar and pSbCl3 = 0.22 bar. Calculate the equi

Lorico [155]

Answer:

2.7 × 10⁻⁴ bar

Explanation:

Let's consider the following reaction at equilibrium.

SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)

The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.

Kp = pSbCl₃ × pCl₂ / pSbCl₅

pCl₂ = Kp × pSbCl₅ / pSbCl₃

pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22

pCl₂ = 2.7 × 10⁻⁴ bar

7 0

3 years ago

0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction

algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce $\frac{2}{1}$ = $\frac{x}{0.50}$

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

$\frac{2}{2}$ = $\frac{x}{0.6}$

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

$\frac{2}{3}$ = $\frac{x}{0.9}$

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.