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hichkok12 [17]
3 years ago
11

Does acid rain make new substances?

Chemistry
1 answer:
ivanzaharov [21]3 years ago
3 0

Answer:

Over the past few decades, humans have released so many different chemicals into the air that they have changed the mix of gases in the atmosphere. ... In addition, the exhaust from cars, trucks, and buses releases nitrogen oxides and sulfur dioxide into the air.

You might be interested in
Given the following equation, what would be the theoretical yield in grams of acetylsalicylic acid (C9H8O4) that can be produced
Mariana [72]

Answer:

The correct answer is 2.6 g C₉H₈O₄

Explanation:

-First we have to write and to balance the chemical equation:

C₇H₆O₃ + C₄H₆O₃ ⇄ C₉H₈O₄ + C₂H₄O₂

-Then, we calculate the molecular masses of reactants and products:

1 mol C₇H₆O₃= (12 g/mol C x 7) + (1 g/mol H x 6) + (16 g/mol O x 3)= 138 g

1 mol C₄H₆O₃= (12 g/mol C x 4) + (1 g/mol H x 6) + (16 g/mol O x 3)= 102 g

1 mol C₉H₈O₄= (12 g/mol C x 9) + (1 g/mol H x 8) + (16 g/mol O x 4)= 180 g

1 mol C₂H₄O₂= (12 g/mol C x 2) + (1 g/mol H x 4) + (16 g/mol O x 2)= 60 g

The mass balance is correct because:

mass reactants = mass products

138 g + 102 g = 180 g + 60 g

     240 g       =     240 g

-Now we use the masses from the chemical equation to calculate how reactant we need. We know that 138 g of salicylic acid (C₇H₆O₃) react with 102 g of acetic anhydride (C₄H₆O₃). So, the grams of acetic anhydride we need to react with 2 g of salicylic acid will be:

138 g C₇H₆O₃------------------- 102 g C₄H₆O₃

2.0 g C₇H₆O₃ -------------------- x= (2.0 x 102)/138 = 1.48 g C₄H₆O₃

If we compare, the amount of C₄H₆O₃ we need (1.48 g) is lesser than the amount we have (8 g), so C₄H₆O₃ is the excess reactant and C₇H₆O₃ is the limiting reactant.

- Finally, <u>we use the limiting reactant</u> to calculate the theoretical yield in grams of C₉H₈O₄. From the chemical equation, we know that 138 g C₇H₆O₃ yield 180 g of C₉H₈O₄. We have 2.0 g, so:

138 g C₇H₆O₃------------------- 180 g C₉H₈O₄

2.0 g C₇H₆O₃ -------------------- x= (2.0 x 180)/138 = 2.6 g C₉H₈O₄

8 0
3 years ago
Read 2 more answers
A mixture of helium and neon gases has a density of 0.2460 g/L at 35.2°C and 288 torr. What is the mole fraction of neon in this
pshichka [43]
We know the volume of one mole of gas at 273 K and 760 Torr is 22.4 L. Using
(PV)/T = constant
We can calculate the volume of the gasses at the given conditions:
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4) / 273 = (288 V₂) / 308.2
V₂ = 66.7 L
Mass of He: 4
Mass of Ne: 20
Fraction of Ne: x
Fraction of He: 1 - x
avg density = (∑(component fraction × component mass))/volume
0.2460 = (20x + 4(1 - x))/ 66.7
x = 0.775
6 0
4 years ago
¿POR QUÉ SE LE LLAMA INDICADOR UNIVERSAL AL PAPEL p H?
maksim [4K]

Answer:

ª

Explanation:

⠄⠄⠄⠄⠄⠄⢴⡶⣶⣶⣶⡒⣶⣶⣖⠢⡄⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⢠⣿⣋⣿⣿⣉⣿⣿⣯⣧⡰⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⣿⣿⣹⣿⣿⣏⣿⣿⡗⣿⣿⠁⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⠟⡛⣉⣭⣭⣭⠌⠛⡻⢿⣿⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⠄⠄⣤⡌⣿⣷⣯⣭⣿⡆⣈⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⠄⢻⣿⣿⣿⣿⣿⣿⣿⣷⢛⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⠄⠄⢻⣷⣽⣿⣿⣿⢿⠃⣼⣧⣀⠄⠄⠄⠄⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⠄⠄⠄⠄⣛⣻⣿⠟⣀⡜⣻⢿⣿⣿⣶⣤⡀⠄⠄⠄⠄⠄

⠄⠄⠄⠄⠄⠄⠄⠄⢠⣤⣀⣨⣥⣾⢟⣧⣿⠸⣿⣿⣿⣿⣿⣤⡀⠄⠄⠄

⠄⠄⠄⠄⠄⠄⠄⠄⢟⣫⣯⡻⣋⣵⣟⡼⣛⠴⣫⣭⣽⣿⣷⣭⡻⣦⡀⠄

⠄⠄⠄⠄⠄⠄⠄⢰⣿⣿⣿⢏⣽⣿⢋⣾⡟⢺⣿⣿⣿⣿⣿⣿⣷⢹⣷⠄

⠄⠄⠄⠄⠄⠄⠄⣿⣿⣿⢣⣿⣿⣿⢸⣿⡇⣾⣿⠏⠉⣿⣿⣿⡇⣿⣿⡆

⠄⠄⠄⠄⠄⠄⠄⣿⣿⣿⢸⣿⣿⣿⠸⣿⡇⣿⣿⡆⣼⣿⣿⣿⡇⣿⣿⡇

⠇⢀⠄⠄⠄⠄⠄⠘⣿⣿⡘⣿⣿⣷⢀⣿⣷⣿⣿⡿⠿⢿⣿⣿⡇⣩⣿⡇

⣿⣿⠃⠄⠄⠄⠄⠄⠄⢻⣷⠙⠛⠋⣿⣿⣿⣿⣿⣷⣶⣿⣿⣿⡇⣿⣿⡇

3 0
3 years ago
All are true for the isomerase reaction of glucose-6-phosphate to fructose-6-phosphate except:
leonid [27]
B it is an aldose to ketose isomerization
4 0
4 years ago
How many milliliters of a 0.640 M solution of KBr would be required to contain 17.2 grams of KBr
mamaluj [8]

Answer:

227 mL KBr

Explanation:

To find the amount of milliliters KBr, you need to (1) convert grams to moles (via molar mass from values on periodic table), then (2) find the amount of liters KBr (via molarity equation using molarity and moles), and then (3) convert liters to milliliters. The final answer should have 3 sig figs to match the amount of sig figs in the given values.

<u>(Step 1)</u>

Molar Mass (KBr): 39.098 g/mol + 79.904 g/mol

Molar Mass (KBr): 119.002 g/mol

17.2 grams KBr            1 mole
-----------------------  x  ------------------  =  0.145 moles KBr
                                  119.002 g

<u>(Step 2)</u>

Molarity (M) = moles / volume (L)

0.640 M = 0.145 moles / volume

(0.640 M) x (volume) = 0.145 moles

volume = (0.145 moles) / (0.640 M)

volume = 0.227 L

<u></u>

<u>(Step 3)</u>

<u></u>

0.227 L KBr         1,000 mL
------------------  x  -----------------  = 227 mL KBr
                                1 L

3 0
2 years ago
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