Heated bricks or blocks of iron were used long ago to warm beds. A 1.49 -kg block of iron heated to 155 Celsius would release how many joules of heat as it cooled to 22 Celsius ?
Answers
Shc of iron (constant)= 0.4494J/³C for 1g
1.49kg=1490g
Q=1490x(22-155)x0.4494
Q=89057.598J
Final answer:
The question pertains to the concept of specific heat capacity. Using the formula 'q = mcΔT' where 'q' is the heat transferred, 'm' is the mass of the substance, 'c' is the specific heat and 'ΔT' is the temperature change, we can calculate how much heat a block of iron would release when it cools.
Explanation:
To solve your question, we need to understand the concept of specific heat capacity, which is an intensive property that depends only on the type of substance absorbing or releasing heat. The specific heat capacity (c) of a substance, commonly called its "specific heat," is the quantity of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. The specific heat of iron is :
0.449 J/g°C
To calculate the amount of heat released, we need to use the formula for heat transfer as follows: q = mcΔT where:
- q is the heat transferred
- m is the mass of the substance
- c is the specific heat of the substance
- ΔT is the change in temperature
In this case, the mass of iron is 1.49 kg or 1490 g, the specific heat capacity of iron is 0.449 J/g°C, and the change in temperature is 155°C - 22°C = 133°C.
By multipying these values in the formula we get: q = 1490g x 0.449 J/g°C x 133°C. Therefore, the block of iron would release calculated amount of Joules of heat as it cooled from 155°C to 22°C.
Learn more about specific heat capacity here:
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Answers
HClO + H2O ⇄ H3O+ +OCl-
Then the equilibrium constant, Ka, of dilute HClO would be:
Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x
Substituting the excess (E) concentration to the Ka equation:
Simplifying the equation would yield a quadratic equation:
The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.
x = 1.688 x 10^-6 M = [H3O+] = [ClO-]
Then, you can determine the conc of [OH-] through pH.
pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M
Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M
Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.
Thus in relative amounts, the order would be
H2O >>> HClO > H3O+ = ClO- > OH-
The Formula is:
Molar Mass of an Element = x Relative mass of atoms Molar mass constant (1g / mol)
Further Explanation
Understand molar mass
Molar mass = mass (in grams) of 1-mol substances
Molar mass is the mass (in grams) of one mole of a substance. By using the atomic mass of an element and multiplying it by the conversion factor of grams per mole (g / mol), you can calculate the molar mass of the element.
Look for the relative atomic mass of elements
Relative atomic mass:
Hydrogen = 1,007
Carbon = 12.0107
Oxygen = 15,9994
Chlorine = 35,453
The relative atomic mass of an element is the average mass of a sample of all its isotopes in atomic units. This information can be found in the periodic table of elements.
Multiply the atomic mass relative to the molar mass constant
Molar Mass of an Element = x Relative mass of atoms Molar mass constant (1g / mol)
Hydrogen = 1,007 x 1g / mol = 1,007 g / mol
Carbon = 12.0107 g / mol
Oxygen = 15,999 g / mol
Chlorine = 35,453 g / mol
H2: 1,007 x 2 = 2,014 g / mol
O2: 15,999 x 2 = 31,9988 g / mol
C2: 35,453 x 2 = 70,096 g / mol
example:
Hydrochloric acid
HCl -> 1 hydrogen atom, 1 chlorine atom
Glucose
C6H12O6 -> 6 carbon atoms, 12 hydrogen atoms, 6 oxygen atoms
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Details
Class: Middle/High School
Subject: Chemistry
Keywords: Mass, Molar, atoms
B._25 grams
C._50 grams
D._100 grams
Answers
Answer : The correct option is, (C) 50 grams
Solution :
Formula used :
where,
a = amount of reactant left after n-half lives = ?
= Initial amount of the reactant = 100 g
n = number of half lives = 1
Putting values in above equation, we get:
Therefore, the amount of uranium-238 remain will be, 50 g
This is because a half life is the time it takes for a sample of an unstable isotope to spontaneously so that half of the original sample remains
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