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3 years ago

A 126-gram sample of titanium metal is heated from 20.0°C to

1 answer:

const2013 [10]3 years ago

3 0

The specific heat for the titanium metal is 0.524 J/g°C. Here, ΔT = T₂ - T₁ = 45.4 - 20 = 25.4°C.
Explanation:
Given,
Q = 1.68 kJ = 1680 Joules
mass = 126 grams
T₁ = 20°C
T₂ = 45.4°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
Here, ΔT = T₂ - T₁ = 45.4 - 20 = 25.4°C.
Substituting values,
1680 = (126)(25.4)(Cp)
By solving,
Cp = 0.524 J/g°C.
The specific heat for the titanium metal is 0.524 J/g°C.

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How do I convert/scale a recipe using a t chart

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Answer:

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Explanation:

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3 years ago

t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol

spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Initial conc. 0.10 0.10 0.50

At eqm. (0.10-x) (0.10-x) (0.50+2x)

As we are given:

$K_c=5.00\times 10^2$

The expression for equilibrium constant is:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

Now put all the given values in this expression, we get:

$5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}$

x = 0.0713 and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M

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4 years ago

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3 years ago

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4 years ago

Which one of the following pairs of acids and conjugate bases is incorrectly labeled or incorrectly matched? Acid Conjugate Base

vekshin1

Answer : The incorrect option is, (d) $NH_4^+/NH_2^-$

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor.

(a) The equilibrium reaction will be,

$HF+H_2O\rightleftharpoons F^-+H_3O^+$

In this reaction, $HF$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms $F^-$ and $H_3O^+$ are conjugate base and acid respectively.

In this reaction, $HF/F^-$ are act as a conjugate acid-base.

(b) The equilibrium reaction will be,

$HClO+H_2O\rightleftharpoons ClO^-+H_3O^+$

In this reaction, $HClO$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms $ClO^-$ and $H_3O^+$ are conjugate base and acid respectively.

In this reaction, $HClO/ClO^-$ are act as a conjugate acid-base.

(c) The equilibrium reaction will be,

$H_2O+H_2O\rightleftharpoons OH^-+H_3O^+$

In this reaction, $H_2O$ is an acid that donate a proton or hydrogen to

In this reaction, $H_2O/OH^-$ are act as a conjugate acid-base.

(d) The equilibrium reaction will be,

$NH_4^++H_2O\rightleftharpoons NH_3+H_3O^+$

In this reaction, $NH_4^+$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms $NH_3$ and $H_3O^+$ are conjugate base and acid respectively.

In this reaction, $NH_4^+/NH_3$ are act as a conjugate acid-base.

(e) The equilibrium reaction will be,

$H_3O^++H_2O\rightleftharpoons H_2O+H_3O^+$

In this reaction, $H_3O^+$ is an acid that donate a proton or hydrogen to $H_2O$ base and it forms

In this reaction, $H_3O^+/H_2O$ are act as a conjugate acid-base.

From this we conclude that that, $NH_4^+/NH_2^-$ pairs of acids and conjugate bases is incorrectly labeled or incorrectly matched.

Hence, the incorrect option is, (d) $NH_4^+/NH_2^-$

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3 years ago