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3 years ago

How much energy is required to raise the temperature of 3 kg of aluminum

Chemistry

1 answer:

FrozenT [24]3 years ago

6 0

Answer:

Explanation:

Q = mcAT= should be mc(T2-T1)

for Aluminum

Q = 3 1000g Al x(0.897J/gC x(23-18 C)= 16722 J

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How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

3 years ago

An aqueous solution is analyzed and found to contain potassium ions and sulfite ions. write the equation for the dissolution of

enot [183]

1) T<span>he dissolution of the salt potassium sulfite:
K</span>₂SO₃(aq) → 2K⁺(aq) + SO₃²⁻(aq).
Potassium has +1 charge because it lost one electron to accomplish stabile electron configuration of noble gas argon.
2) From dissolution reaction: n(K⁺) : n(SO₃²⁻) = 2 : 1.
n(K⁺) = 0.700 mol.
0.700 mol : n(SO₃²⁻) = 2 : 1.
n(SO₃²⁻) = 0.700 mol ÷ 2.
n(SO₃²⁻) = 0.350 mol; amount of sulfite anions.

7 0

3 years ago

3. The dry air in the Alps helped to prevent damage to the cells in the Iceman's body.

Sever21 [200]

Answer:

False

Explanation:

because they do not prevent damage they make damage

3 0

3 years ago

What happens to the liquid in a thermometer when it is moved from cold water to boiling water?

MissTica

Nothing it just changes temp as it is protected by glass layers

5 0

3 years ago

Read 2 more answers

Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to th

Luba_88 [7]

Answer:

The percent yield of the reaction is 82%

Explanation:

First step: make the chemist equation.

2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)

As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3

Second step: Find out the moles in the reactant.

Molar weight Fe2O3: 159.7 g/m

Mass / Molar weight = moles

50 g /159.7 g/m = 0.313 moles

Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.

1 mol Fe2O3 ____ 2Fe

0.313 mol Fe2O3 ____ 0.626 moles

Molar weight Fe = 55.85 g/m

Moles . molar weight = mass

55.85g/m . 0.626m = 34.9 grams

This will be the 100% yield of the reaction but we only made 28.65 g

34.9 g ____ 100%

28.65 g ____ 82.09 %

3 0

3 years ago

Read 2 more answers