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Rasek [7]
3 years ago
15

What are the signs that a chemical reaction has occurred

Chemistry
2 answers:
Finger [1]3 years ago
8 0
Well. There is the color change and the formation of bubbles. 
MrRa [10]3 years ago
8 0
Color change, formation of a precipitate, heat, formation of a gas, and odor
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Ammonium Iodide dissociates reversibly to ammonia and hydrogen iodide:
NeTakaya

Answer:

The partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C Is 0.103 atm.

The correct option is A.

Explanation;

NH4I(s) ⇋ NH3(g) + HI(g)Kp = 0.215 at 400°C

NH4I(s)= 0.215

NH3(g)=0.103

HI(g)Kp=0.112

Therefore = 0.103 +0.112= 0.215

Therefore the partial pressure of ammonia at equilibrium is 0.103 atm

7 0
3 years ago
What does this mean???
satela [25.4K]

Answer:

Its known as static electricity

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2 years ago
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3 years ago
From where do nuclear power plants get uranium?
Anvisha [2.4K]

Answer:

Uranium mines operate in many countries, but more than 85% of uranium is produced in six countries: Kazakhstan, Canada, Australia, Namibia, Niger, and Russia. Historically, conventional mines open pit or underground were the main source of uranium.

Explanation:

Hope this helps

4 0
2 years ago
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug
OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 =  log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] <em>(1)</em>

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = <em>0.035M</em>

Thus: [CH₃COO⁻] = 0.100M - 0.035M = <em>0.065M</em>

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles +  2.646x10⁻³ moles = <em>8.071x10⁻³ moles of CH₃COOH</em>. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = <em>0.296</em>

4 0
3 years ago
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