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Lelechka [254]
3 years ago
13

Based on the chart, determine the identity of a substance that increases in temperature by 11.4oC when 1250J of energy are added

to a 55g sample.
Substance Specific Heat (J/g oC)
Lithium 3.56
Vegetable Oil 1.99
Air 1.02
Iron 0.444
Sand 0.290
Gold 0.129

Salt

Air

Lithium

Vegetable Oil
Chemistry
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

LIthium vegetable oil

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6 0
3 years ago
A student decomposed 3.67g of copper (ii) hydroxide into copper (ii) oxide. how many ml of 3m h2so4 is need to react with all th
vredina [299]
In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide. 
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂  - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄ 
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present 
so if 3 mol are present in 1000 ml 
then volume for 0.038 mol = 1000/3 * 0.038 
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6 0
4 years ago
2. What is the mass of 5.3 moles of CaCl2?
Tanya [424]

Answer:

590 g CaCl₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

5.3 mol CaCl₂

<u>Step 2: Identify Conversions</u>

Molar Mass of Ca - 40.08 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of CaCl₂ - 40.08 + 2(35.45) = 110.98 g/mol

<u>Step 3: Convert</u>

<u />5.3 \ mol \ CaCl_2(\frac{110.98 \ g \ CaCl_2}{1 \ mol \ CaCl_2} ) = 588.194 g CaCl₂

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules and round.</em>

588.194 g CaCl₂ ≈ 590 g CaCl₂

5 0
3 years ago
A compound decomposes by a first-order process. What is the half-life of the compound if 25.0% of the compound decomposes in 60.
amid [387]

Answer : The half-life of the compound is, 145 years.

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time passed by the sample  = 60.0 min

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 25 = 75 g

Now put all the given values in above equation, we get

k=\frac{2.303}{60.0}\log\frac{100g}{75g}

k=4.79\times 10^{-3}\text{ years}^{-1}

Now we have to calculate the half-life of the compound.

k=\frac{0.693}{t_{1/2}}

4.79\times 10^{-3}\text{ years}^{-1}=\frac{0.693}{t_{1/2}}

t_{1/2}=144.676\text{ years}\approx 145\text{ years}

Therefore, the half-life of the compound is, 145 years.

8 0
4 years ago
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