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Novosadov [1.4K]
3 years ago
9

A closed container of N2O4 and NO2 is at equilibrium. NO2 is added (increasing the concentration) to the container.

Chemistry
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

closed container of N2O4 and NO2 is at equilibrium. NO2 is added (increasing the concentration) to the container.

N2O4 (g)

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Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg
FrozenT [24]

Answer:

350 g dye

0.705 mol

2.9 × 10⁴ L

Explanation:

The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:

70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye

The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:

350 g × (1 mol / 496.42 g) = 0.705 mol

The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:

3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L

8 0
3 years ago
Given 0.02 of KF, how many liters of solution are needed to make a 7.2 solution?
Dmitry_Shevchenko [17]

Answer:

1/360

Explanation:

let x = liters

molarity=moles of solute/liters of solution, 7.2=0.02/x or 7.2=(1/50)(1/x), 7.2(50)=(1/x), 360(x)=1, x=1/360

7 0
3 years ago
Melissa is interested in her family tree and how her family has changed over its many generations. Melissa probably more closely
tresset_1 [31]

Answer: A Confucian

Explanation:

In the year 2005, the Guinness Book of World Records recognized the Confucius genealogical line as the longest family tree in history. They have a record of 86 generations over 2,500 years. Confucius (551 to 479 BCE) is believed to have about 3 million descendants all over the world.

So the interest Melissa has in her family tree and also wanting to know how her family has changed over its many generations probably suggests that Melissa more closely resembles a Confucian.

6 0
3 years ago
Read 2 more answers
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 solution is made by adding 29.1 mL of concentrated perchloric acid ( 70.5 wt% , density 1.67 g/mL ) to some water in a volumet
Lera25 [3.4K]

Answer:

The concentration of the solution is 1.364 molar.

Explanation:

Volume of perchloric acid = 29.1 mL

Mass of the solution = m

Density of the solution = 1.67 g/mL

m=1.67 g/mL\times 29.1 mL=48.597 g

Percentage of perchloric acid in 48.597 solution :70.5 %

Mass of perchloric acid in 48.597 solution :

= \frac{70.5}{100}\times 48.597 = 34.261 g

Moles of perchloric acid = \frac{34.261 g}{100.46 g/mol}=0.3410 mol

In 29.1 mL of solution water is added and volume was changed to 250 mL.

So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)

Molarity=\frac{Moles}{Volume (L)}

=\frac{0.3410 mol}{0.250 L}=1.364 M

The concentration of the solution is 1.364 molar.

6 0
3 years ago
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