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Step2247 [10]
3 years ago
11

A cool, yellow-orange flame is used to heat the crucible. Would this affect the mass of the crucible? If so, how?

Chemistry
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer:

yes

Explanation:

Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.

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Name the cycloalkanes with molecular formula c6h12 that have a 3-membered ring and two substituents.
Tanzania [10]

Answer:

1-ethyl-2-methyl cyclopropane.

Explanation:

  • The structure of the molecule will be as shown in the attached image.
  • The molecular formula of the compound is C₆H₁₂.
  • It has 3 membered ring with 3 C atoms and two substituents one of them with one C atom (methyl) and the other with 2 C atoms (ethyl).
  • The ring consist of 3 C atoms, so its name is cyclo propane.
  • We numbering the atoms of the ring that give the ethyl substituent the low no. (1) and then methyl group take no. (2).
  • <em>Thus, the name of the compound is 1-ethyl-2-methyl cyclopropane.</em>

3 0
3 years ago
If 0.500 mol of each of the following solutes is dissolved in 2.0 L of water, which will cause the greatest increase in the boil
serious [3.7K]

Answer is: D. Na2SO4.

b(solution) = 0.500 mol ÷ 2.0 L.

b(solution) = 0.250 mol/L.

b(solution) = 0.250 m; molality of the solutions.

ΔT = Kf · b(solution) · i.

Kf - the freezing point depression constant.

i - Van 't Hoff factor.

Dissociation of sodium sulfate in water: Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq).

Sodium sulfate dissociates on sodium cations and sulfate anion, sodium sulfate has approximately i = 3.

Sodium chloride (NaCl) and potassium iodide (KI) have Van 't Hoff factor approximately i = 2.

Carbon dioxide (CO₂) has covalent bonds (i = 1, do not dissociate on ions).

Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.

8 0
3 years ago
Read 2 more answers
3. Sexual reproduction is the process of creating a new organism by combining the
weqwewe [10]

Answer:

chromosomes is responsible

6 0
2 years ago
Proton-alpha: "One time I lived in an environment where next door lived identical twins, Proton-beta-1 and Proton-beta-2. Boy, d
Mashcka [7]

Answer:

Explanation:

C) What is the multiplicity of Proton-alpha's signal in this scenario when there are 2 identical protons "next door"?

Based on n+1 rule. Here n=2 (identical beta protons).

2+1=3

So the multiplicity of alpha proton is triplet, .

D) For molecules containing only single bonds (we'll discuss the influence of double bonds in a future lecture), what is the adjective that describes the position of protons that split a "next door neighbor's" signal?

    The meaning of the adjective is this: the multiplicity of beta protons is singlet only (no spliting) in absence of alpha proton . But beta protons splits as doublet (n=1) in the presence of alpha proton,

E) How many bonds connect these "splitting next door neighbors"?

There are 3 bonds in between alpha and beta protons in a molecule.

F) What is the multiplicity of the Proton-betas' signal?

Following the  n+1 rule, here n=1 (1 alpha proton) so 1+1=2. Hence it is  a doublet.

7 0
3 years ago
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -&gt; 2Al2O3 In a certain experiment, 4.6g Al was react
stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

<h3>78.2% </h3>
8 0
3 years ago
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