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2 years ago

13

Three hundred grams of aluminum are heated from −50 ◦c to 300 ◦c. if the specific heat of aluminum is 0.88 j/(g ◦c), what is the

change in entropy of the aluminum?

1 answer:

enyata [817]2 years ago

4 0

Q = mCΔT
Q is heat in Joules, m is mass, C is specific heat, and ΔT is change in temp

Q = (300g)(.88)(350 degrees) = 92400 J or 92.4 kJ

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Which phase change is exothermic?

DerKrebs [107]

Freezing, condensation, Deposition.

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2 years ago

If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the

Maurinko [17]

Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

$w/w\% = \frac{x}{m}\times 100$

$3\%=\frac{x}{5.125 g}\times 100$

$x=\frac{3\times 5.125 g}{100}=0.15375 g$

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

$=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol$

0.004522 moles of hydrogen peroxide molecules are present.

5 0

2 years ago

4. Write the electronic configuration of first 20 elements in the periodic table.

DIA [1.3K]

Answer:

Name Atomic Number Electron Configuration Period 1 Hydrogen 1 1s1 Helium 2 1s2 Period 2 Lithium 3 1s2 2s1 Beryllium 4 1s2 2s2 Boron 5 1s2 2s22p1 Carbon 6 1s2 2s22p2 Nitrogen 7 1s2 2s22p3 Oxygen 8 1s2 2s22p4 Fluorine 9 1s2 2s22p5 Neon 10 1s2 2s22p6 Period 3 Sodium 11 1s2 2s22p63s1 Magnesium 12 1s2 2s22p63s2 Aluminum 13

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2 years ago

Which of the following elements could be transmutated into silver by bombardment with a positron?

Elena L [17]

Answer:

I think it's gold but I'm not sure, sorry

Explanation:

good luck tho :)

6 0

2 years ago

If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th

qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

$K_c=4.74\times 10^4$

Concentration at equilibrium:

$[COBr_2]=1.58\times 10^{-6}M$

$[Co]=2.78\times 10^{-3}M$

$[Br_2]=2.51\times 10^{-5}M$

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

The expression for reaction quotient will be :

$Q=\frac{[CO][Br_2]}{[COBr_2]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

$Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}$

The given equilibrium constant value is, $K_c=4.74\times 10^4$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K_c$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the $Q$ that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0

2 years ago