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Ghella [55]
3 years ago
12

How should you add metal to the graduated cylinder so that it will not crack the glass at the bottom?

Chemistry
1 answer:
ella [17]3 years ago
5 0

Answer:

you tilt the cylinder at a slight angle so that the metal slides down the sides, rather than drops all it`s weight to the bottom

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Never mind wrong question
Leokris [45]
Its okay my friend. you dont need to over stress it.
7 0
3 years ago
Read 2 more answers
Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation:
weqwewe [10]

Answer:

D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M

Explanation:

Based on the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

And knowing:

Kc = [PCl₃] [Cl₂] / [PCl₅] = 1.80

When you add PCl₅ into a flask, this gas will react producing PCl₃ and Cl₂ until [PCl₃] [Cl₂] / [PCl₅] = 1.80

This could be written as:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

<em>Where X represents the moles of PCl₅ that react, </em><em>reaction coordinate.</em>

Replacing in Kc expression:

[PCl₃] [Cl₂] / [PCl₅] = 1.80

[X [X] / [0.125 - X] = 1.80

X² = 0.225 - 1.80X

0 = -X² -1.80X + 0.225

Solving for X:

X = -1.9M → False solution, there is no negative concentrations

X = 0.11735M → Right solution.

Replacing, concentrations in equilibrium are:

[PCl₃] = X

[Cl₂] = X

[PCl₅] = 0.125M - X

[PCl₃] = 0.117M

[Cl₂] = 0.117M

[PCl₅] = 0.00765M

And right option is:

<h3>D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M</h3>
7 0
3 years ago
Describe any new ways you can change the polarity of the three-atom molecule.
yan [13]

-  By changing three atom bond angle , we can change molecular polarity. if bond angle is 120° and all atom has same electronegativity, the resultant polarity ( dipole moment ) become zero. we can change bond anngle either less than or greater than 120°, but not exactly 120

-  Replacing one or more atom with different atoms having electro negativity values also make molecular poles.

-  By placing molecules under external electric field or magnetic field also causes to the molecule.

5 0
3 years ago
Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation o
Korolek [52]

The rate of disappearance of chlorine gas : 0.2 mol/dm³

<h3>Further explanation</h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

\tt aA+bB\rightarrow cC+dD

The rate reaction :

\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}

Reaction for formation CCl₄ :

<em>CH₄+4Cl₂⇒CCl₄+4HCl</em>

<em />

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of  CCl₄  = reaction rate x coefficient of  CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of  Cl₂  = reaction rate x coefficient of  Cl₂

Rate of disappearance of  Cl₂ = 0.05 x 4 = 0.2 mol/dm³

4 0
3 years ago
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