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nirvana33 [79]
3 years ago
8

How to do 3 and 4 questions

Chemistry
1 answer:
Kipish [7]3 years ago
5 0

Answer:

3) Q = -836.8 J.

4) Q = 950J.

Explanation:

Hello there!

In this case, for those calorimetry problems, we use the general equation:

Q=mC\Delta T

Thus, we proceed as follows:

3) Here, the temperature difference is from 80 °C to 40 °C, the mass is 5.0 g and the specific heat 4.184 in SI units; thus, the heat is calculated as follows:

Q=(5g)(4.184\frac{J}{g\°C} )(40\°C-80\°C)\\\\Q=-836.8J

4) Here, the temperature difference is from 100 °C to 200 °C, the mass is 5.0 g and the specific heat about 1.90 in SI units; thus, the heat is calculated as follows:

Q=(5g)(1.90\frac{J}{g\°C} )(200\°C-100\°C)\\\\Q=950J

Regards!

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Answer:

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Explanation:

a) The probability of a red ball being drawn is 12/38, or in a simplified fraction, 6/19. To find the probability that 3 are red you would multiply the probability of the fraction for each, except subtracting one from the total each time as the drawn is done without replacement. This is done as follows: 6/19 × 6/18 × 6/17= 12/323

b) The probability of drawing a blue ball is 8/38, or 4/19. To find that the first one is blue and the rest are red, the equation is done as follows: 4/19 × 6/18 × 6/17 = 8/233

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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.

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<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)

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3 years ago
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