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9966 [12]
3 years ago
7

tate whether the following changes are physical or chemical for rancidipication fixation of water 2 tearing of paper 3 rusting o

f iron 4 electrolysis of water​
Chemistry
1 answer:
damaskus [11]3 years ago
3 0

Answer: Physical change : tearing of paper, fixing of wtaer

Chemical change:  rusting of iron ,  electrolysis of water​, Rancidification

Explanation:

Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.

Example:  tearing of paper, fixing of wtaer

Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.

Example: rusting of iron ,  electrolysis of water​, Rancidification

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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

5 0
3 years ago
A natural disaster caused an ecosystem to experience a significant drop in species richness. What can you conclude about the eco
KATRIN_1 [288]
C. Theres decreased Biodiversity Most likely because of natural or human causes (deforistation)
3 0
3 years ago
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If 16.5g of c6h14o2 are reacted vwith .499 mol of o2. How many moles of co2 should be produced
Fed [463]

Answer:

moles of carbon dioxide produced are 410.9 mol.

Explanation:

Given data:

Mass of C₆H₁₄O₂ = 16.5 g

Moles of O₂ = 499 mol

Moles of CO₂ = ?

First of all we will write the balance chemical equation.

2C₆H₁₄O₂  +  17O₂  →   14CO₂  +  12H₂O

moles of C₆H₁₄O₂  = mass × molar mass

moles of C₆H₁₄O₂ =  16.5 g × 118 g/mol

moles of C₆H₁₄O₂ = 1947 mol

Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.

                 O₂      :     CO₂

                 17       :      14

                499     :      14/17× 499 = 410.9 moles

          C₆H₁₄O₂   :   CO₂

                    2    :      14

                 1947 :     14/2× 1947 =  13629 moles

Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.

5 0
4 years ago
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Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen
andrew-mc [135]
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the limiting reactant in this reaction is C₄H₁₀  This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
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3 years ago
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Answer: i think it is true.
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