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zmey [24]
3 years ago
12

1. Mg + N2 —> MgN2

Chemistry
1 answer:
quester [9]3 years ago
8 0

(1) IS THE BALANCED EQUATION AND REACTANT-> MG AND N2 AND PRODUCT ->MGN2.

(2) IS NOT BALANCED AND REACTANT->CF4 AND BR AND PRODUCT IS CBR4 AND F2

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The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride.
Vlad1618 [11]
5F2 + 2NH3 --> N2F4 + 6HF 
<span>60.1g NH3 / 17g/mole = 3.54moles NH3 </span>
<span>3.54moles NH3 x (5 F2 / 2NH3) x 38g/mole = 335.85g required </span>

<span>5.25g HF / 20g/mole = 0.262moles HF </span>
<span>0.262moles HF x (2NH3 / 6HF) x 17g/mole = 1.49g required </span>

<span>209g / 38g/mole = 5.5moles F2 </span>
<span>5.5moles F2 (1 N2F4 / 5F2) x 66g/mole = 72.6g produced </span>


<span>Li3N + 3H2O --> NH3 + 3LiOH </span>
<span>(37.7g / 34.7g/mole) x (3H2O / 1 Li3N) x 18g/mole = 58.67g required </span>

<span>1.08moles Li3N (1NH3 / 1Li3N) x 6.022x10^23molecules/mole = 6.54x10^23 molecules </span>

<span>10.3L at STP: 10.3L / 22.4L/mole = 0.46moles NH3 produced </span>
<span>0.46moles NH3 x (1Li3N / 1NH3) x 34.7g/mole = 15.96g</span>
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2 years ago
Which type of bonds do polar covalent bonds break down in chemical reactions?
Vlada [557]

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(C) im pretty sure is the answer

Explanation:

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3 years ago
What describes a phase change
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A phase change is when matter changes to from one state ( solid, liquid, gas, plasma) to another.
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If you are running at top speed and try to stop suddenly, what happens?
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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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