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lilavasa [31]
2 years ago
14

“Kids and grown-ups love it so, the happy world of HARIBO!” Who does not like the dark brown liquorice and its lovely taste?! Ta

sty above all is the liquorice-plant. As another ingredient “ammonium chloratum” very often is used as an acidifier. This name is apothecary’s jargon and the old chemical name of ammonium chloride (NH4Cl).
Ammonium chloride, also known as “salt of salmiak”, is allowed up to 20 g/kg in liquorice by the nutrition laws.

Ammonium chloride can be produced from ammonia (NH3) and hydrogen chloride (HCl). You are a student trainee with HARIBO’s. Suddenly the machine in which the ammonia and the hydrogen chloride are combined and added to the liquorice sounds the alarm, but your boss is in his office on an urgent telephone call and does not want to be interrupted. The valves controlling the doses of ammonia and hydrogen chloride have to be adjusted. You have to act quickly. It is very important not to overdose the ammonium chloride, because its amount in the liquorice is controlled regularly by an independent supervision agency.

The machine produces ammonium chloride for a thousand bags of 200 g liquorice each.
Chemistry
1 answer:
KiRa [710]2 years ago
3 0

Answer:

Any Questions?

So i can give you the answer

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katrin [286]

Answer:

CO(g) + 2H₂(g) → CH₃OH(l)

Explanation:

Carbon monoxide has molecular formula CO, molecular hydrogen has formula H₂, and methanol is CH₃OH.

The reactants are CO and H₂ and the product CH₃OH:

CO(g) + H₂(g) → CH₃OH(l)

To balance the equation, the elements must have the same amount on each side. C and O are balanced, but there is 4H in the product and only 2 in the reactant, so we multiply H₂ for 2:

CO(g) + 2H₂(g) → CH₃OH(l)

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3 years ago
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Nezavi [6.7K]

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I hope that the right answer.

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Find the number of CoCl2 units present in a 0.78 mol sample.
stich3 [128]

\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

1 mole of \sf{ COCl_2} has 6.022 × 10²³ molecules of the given compound.

So, 0.78 mole of \sf{COCl_2} will have ~

\qquad \sf  \dashrightarrow \: 0.78 \times 6.022 \times 10 {}^{23}

\qquad \sf  \dashrightarrow \:  \approx4.7 \times 10 {}^{23}  \:  \:  \: molecules

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Answer:

true

Explanation:

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