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3 years ago

We apply the same amount of energy to 10.0-g samples of aluminum, iron, and silver which begin at the same temperature.

Chemistry

1 answer:

maw [93]3 years ago

7 0

Answer:

Explanation:

A sound knowledge of specific heat capacity of the metals is required in this case.

The specific heat capacity of a metal is the quantity of heat required to the raise the temperature of a unit mass of it by 1°C.

It is related to quantity of heat using the expression below;

H = m c Δt

where m is the mass

c is the specific heat capacity

Δt is the temperature change

let us make the specific the subject of the expression;

c = $\frac{H}{m x change in temperature}$

we can see that there is an inverse relationship between specific heat and temperature change.

The specific heat capacity of a body is an intensive property that is unique to the metal.

The higher the specific heat capacity, the lower the amount of temperature change in it.

Let us find the specific heat capacity of the given metals;

Aluminium 0.897J/gK

Iron 0.412J/gK

Silver 0.24J/gK

After the heat is supplied,

Silver > Iron > Aluminium in terms of temperature change

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Name and describe three measures of central tendency used to summarize data.

Yuliya22 [10]

The answer is mean,mode and median

6 0

4 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

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

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

4 0

3 years ago

14.2 grams of Na2SO4 is dissolved in water to make a 2.50 L

Trava [24]

Answer:

0.04 M

Explanation:

Given data:

Mass of Na₂SO₄= 14.2 g

Volume of solution = 2.50 L

Molarity of solution = ?

Solution:

Number of moles of Na₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 14.2 g/ 142.04 g/mol

Number of moles = 0.1 mol

Molarity :

Molarity = number of moles of solute / volume of solution in L

Molarity = 0.1 mol / 2.50 L

Molarity = 0.04 M

6 0

3 years ago

How many reactant molecules and product gas molecules are in this equation?

Mice21 [21]

Answer:

N₂ = 6.022 × 10²³ molecules

H₂ = 18.066 × 10²³ molecules

NH₃ = 12.044 × 10²³ molecules

Explanation:

Chemical equation;

N₂ + 3H₂ → 2NH₃

It can be seen that there are one mole of nitrogen three mole of hydrogen and two moles of ammonia are present in this equation. The number of molecules of reactant and product would be calculated by using Avogadro number.

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

Number of molecules of nitrogen gas:

1 mol = 6.022 × 10²³ molecules

Number of molecules of hydrogen:

3 mol × 6.022 × 10²³ molecules/ 1 mol

18.066 × 10²³ molecules

Number of molecules of ammonia:

2 mol × 6.022 × 10²³ molecules/ 1 mol

12.044 × 10²³ molecules

6 0

4 years ago

Chemistry is defined as:

attashe74 [19]

Answer:

Eeeeeeeeeeee

Explanation:

mark me as a brainlist

3 0

3 years ago