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ivanzaharov [21]
3 years ago
12

Two easy uses of mixture​

Chemistry
2 answers:
padilas [110]3 years ago
5 0

Answer:

Explanation:

Here are a few more examples:

1. Smoke and fog (Smog)

2. Dirt and water (Mud)

lyudmila [28]3 years ago
4 0

Explanation:

it helps to make juices.

It helps to make concentrated acid into dilute acid.

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An organic compound composed of carbon and hydrogen connected only by single bonds is an ________. alkane alkene alkyne aromatic
My name is Ann [436]
Alkane, the others have double covalent bonds
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3 years ago
Rock transfers heat rapidly.truefalse
MaRussiya [10]
No, it does not. I believe it's false. 
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3 years ago
Is the electrochemical cell spontaneous or not spontaneous as written at 25 ∘C ? spontaneous not spontaneous Calculate the poten
givi [52]

Answer:

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

Explanation:

Let's consider the oxidation and reduction half-reactions and the global reaction.

Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻

Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)

Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red,cat - E°red,an

E° = 0.771 V - 0.154 V = 0.617 V

The Nernst equation allows us to calculate the cell potential (E) under the given conditions.

E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

4 0
3 years ago
Which type of magma would you expect to produce the most explosive volcanic eruption?
photoshop1234 [79]

answer: rhyolitic magma

explanation: rhyolitic magma is responsible for most violent volcanic eruptions. rhyolitic magma erupts catastrophically with high intensity gas content.

- cherry :)

4 0
3 years ago
How many grams of sodium sulfide are formed if 1.80 g of hydrogen sulfide is bubbled into a solution containing 2.40 g of sodium
Nookie1986 [14]

The reaction of sodium sulfide with hydrogen sulfide can be shown as

H_2S + 2 NaOH = Na_2S + 2 H_2O \\ Molar mass of H_2S = 34 g /mol\\ Mole of H_2S  = mass /molar mass\\  = 1.8 /34 = 0.051 mole\\ Mole of NaOH  = 2 * 0.051 = 0.102 mole\\ Theoritical mole of NaOH = mass /molar mass\\ = 2.4 /40 = 0.06 mole

Sodium hydroxide is completely reacting and thus act as limiting reactant

Hence, mole of Na_2S required  = 0.06 /2 = 0.03 mole\\ Mass of Na_2S = 0.03 * 78 = 2.31 g

Thus, 2.31 g Na2S produced.

4 0
3 years ago
Read 2 more answers
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