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Black_prince [1.1K]
3 years ago
10

What’s the general form of double replacement reaction?

Chemistry
1 answer:
kondor19780726 [428]3 years ago
4 0

Answer:

neutralization, precipitation, and/or gas

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A gas has a volume of 5.0 L at a pressure of 50 KPa. What happens to the volume when the pressure is increased to 125?
Alexeev081 [22]
The volume becomes two. You have to use the equation P1 x V1 = P2 x V2 
P is pressure and V is volume.
P1 = 50     P2 = 125
V1 = 5       V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.

7 0
3 years ago
ompare your TLC analysis with the results of the column chromatography. More polar molecules move more slowly through silica tha
Assoli18 [71]

Biphenyl will have a higher R value than the Methyl Orange.

Explanation:

Biphenyl is a aromatic hydrocarbon and it is a nonpolar molecule.

Methyl Orange is a organic compound with a -SO₃⁻Na⁺ polar functional group which will induce a high polarity in the compound.

You may find the chemical structures of both molecules in the attached picture.

Column chromatography, which use as stationary phase silica gel, is a good technique for separation of the Methyl Orange from Biphenyl.

Being a non-polar molecule,  Biphenyl will have a higher R value than the Methyl Orange.

To separate them you use a appropriate solvent as eluent, as exemple chloroform, and Biphenyl will elute first from the column and after that, as a separate phase, Methyl Orange will elute thus separating them.

Learn more about:

chromatography

brainly.com/question/10296715

#learnwithBrainly

7 0
3 years ago
The volume of a given mass of agas is 360cm cubic at 50 degrees and 700 milimetres Hg .find it's volume at s.t.p​
saveliy_v [14]

Answer:

V₂  =279.9 cm³

Explanation:

Given data:

Initial volume = 360 cm³

Initial temperature = 50°C

Initial pressure = 700 mmHg

Final volume = ?

Final temperature = 273 k

Final pressure = 1 atm

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Solution:

<em>We will convert the mmHg to atm.</em>

700/760 = 0.92 atm

<em>and °C to kelvin.</em>

50+273 = 323 K

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 0.92 atm × 360 cm³ × 273 K / 323 K ×1 atm

V₂  =  290417.6 atm .cm³.  K  / 323 k. atm

V₂  =279.9 cm³

8 0
4 years ago
Which statements describe an element? Check all that apply.
Molodets [167]

Answer:

<em>Hello, Your answer will be B,C,E, And F.</em>

Explanation:

An element is a substance whose atoms have same number of protons which means all atoms in an element have same atomic number. Elements are simplest substances which cannot be broken down by physical changes or chemical reactions. Elements are classified by their names and symbols. Hope That Helps!

8 0
4 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
4 years ago
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