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gladu [14]
3 years ago
9

It’s is the same one from my recent ones

Chemistry
1 answer:
UkoKoshka [18]3 years ago
5 0
<h2>9. Given : P + O2 = P2O5</h2>

Balanced Chemical Equation: <u>4</u>P + <u>5</u>O2 = <u>2</u>P2O5

<u>Reactant Product</u>

P = (1) 4 | P = (2) 4

O = (2) 10 | O = (5) 10

Reaction Information

Phosphorus + Dioxygen = <u>Phosphorus Pentoxide</u>

Reaction Type: <u>Synthesis</u>

<h2>10. Given: HCl + NaOH = NaCl + H2O </h2>

<u>Reactant Product</u>

H = 2 | H = 2

Cl = 1 | Cl = 1

Na = 1 | Na = 1

O = 1 | O = 1

<h3>Chemical Equation is already in balance.</h3>

Reaction Information<em>:</em><em> </em><em>Hydrogen Chloride </em>+ <em>Sodium Hydroxide</em> = <em>Sodium Chloride </em>+ <em>Water</em>

Reaction Type<u> </u><u>:</u><u> </u><u>Double Displacement (Acid-Base)</u>

<h3>For more info:</h3>

<u>How to balance chemical equation?</u>

brainly.com/question/15939935

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If 30 ml of water containing 15 mg of salicylate is extracted with 30 ml of chloroform, 2.59 mg of salicylate remains in the wat
In-s [12.5K]

Answer:

Distribution coefficient: 4.79

Explanation:

Distribution coefficient is the ratio between equilibrium concentration of non-aqueous phase and aqueous phase where both solvents are inmiscible. The equation for the problem is:

Distribution coefficient: Concentration in chloroform / Concentration in Water

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<em>Concentration in chloroform: (15mg-2.59mg) / 30mL = 0.4137mg/mL</em>

<em />

Distribution coefficient: 0.4137mg/mL / 0.08633mg/mL

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8 0
2 years ago
Could someone help me with this?
Varvara68 [4.7K]

Solving part-1 only

#1

KMnO_4

  • Transition metal is Manganese (Mn)

#2

Actually it's the oxidation number of Mn

Let's find how?

\\ \tt\Rrightarrow x+1+4(-2)=0

\\ \tt\Rrightarrow x+1-8=0

\\ \tt\Rrightarrow x-7=0

\\ \tt\Rrightarrow x=+7

  • x is the oxidation number

#3

  • Purple as per the color of potassium permanganate

#4

\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}

7 0
2 years ago
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