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4 years ago

9

If 5,800 j of energy are applied to a 15.2 kg piece of lead, by how much does the temp change if the specific heat of lead is 0.

128

1 answer:

scoray [572]4 years ago

7 0

Answer:

The answer will be 2.98K

Explanation:

Using the formula:

Q = mc∆T

Q= 5,800 (heat in joules)

m= convert 15.2kg to g which is 15200g (mass in grams)

c= 0.128 J/g °c (Specific heat capacity)

∆T= what we need to find (temperature change)

5800J = 15200g x 0.128 x ∆T

= 2.98K

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A 10-gram sample of zinc loses 560 J of heat and has a final temperature of 100

grandymaker [24]

<h3>Answer:</h3>

Initial temperature is 243.59°C

<h3>Explanation:</h3>

The quantity of heat is calculated by multiplying the mass of a substance by its specific heat capacity and change in temperature.

That is; Q = m×c×ΔT

In this case;

Quantity of heat = 560 J

Mass of the Sample of Zinc = 10 g

Final temperature = 100°C

We are required to determine the initial temperature;

This can be done by replacing the known variables in the formula of finding quantity of heat,

Specific heat capacity, c, of Zinc = 0.39 J/g.°C

Therefore,

560 J = 10 g × 0.39 J/g°C × ΔT

ΔT = 560 J ÷ (3.9 J/°C)

= 143.59°C

But, since the sample of Zinc lost heat then the temperature change will have a negative value.

ΔT = -143.59°C

Then,

ΔT = T(final) - T(initial)

Therefore,

T(initial) = T(final) - ΔT

= 100°C - (-143.59°C)

= 243.59°C

Hence, the initial temperature of zinc sample is 243.59°C

5 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

<u>Part (a) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

<u>Part (b) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago

Sodium metal is quiet soft whereas sodium chloride crystals are quiet hard explain why

stiks02 [169]

In mineralogy and crystallography, a crystal structure<span>is a unique arrangement of atoms in a </span>crystal. Acrystal structure<span> is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice.
Crystals create a harder more fitting structure so they tend to be a lot stronger than other compounds or elements</span>

8 0

4 years ago

What is the difference between atomic elements and molecular elements?

mylen [45]

The molecular element describes the amount of protons, neutrons, and electrons found in an atom

An atomic element is the subject in the formula such as ... sulfur, carbon, or oxygen

6 0

3 years ago

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When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory, what hybrid orbitals are used by

Vlad1618 [11]

Answer:

sp³

Explanation:

Number of hybrid orbitals = ( V + S - C + A ) / 2

Where

H is the number of hybrid orbitals

V is the valence electrons of the central atom = 5

S is the number of single valency atoms = 4

C is the number of cations = 1

A is the number of anions = 0

For PCl₄⁺

Applying the values, we get:

H = ( 5+4-1+0) / 2

= 4

<u>This corresponds to sp³ hybridization.</u>

5 0

3 years ago

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