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Allushta [10]
3 years ago
11

1. Balance the chemical reaction for the production of hydrogen cyanide: __ NH3(g) + __ O2(g) + __ CH4(g) → __ HCN(g) + __ H2O(g

)
(Hint: none of the coefficients are larger than 6)

2.Using the balanced equation in question 1, if 25.1 g of ammonia are reacted, how many grams of hydrogen cyanide will be produced
Chemistry
1 answer:
Solnce55 [7]3 years ago
5 0
<span>2 NH</span>₃<span> + 3 O</span>₂<span> + 2 CH</span>₄<span> </span>⇒<span> 2 HCN + 6 H</span>₂<span>O
2mol                     :           2mol
34g                       :           54g
25,1g                    :           x

x = (25,1g * 54g) / 34g </span>≈ 39,9g<span>
</span>
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What happens to a charged particle that enters a magnetic field?
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Can someone help on this?
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8 0
4 years ago
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol
zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = 10^{-1.9}

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

6 0
3 years ago
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