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3 years ago

The first pesticides relied on natural chemicals, such as those taken from chrysanthemum flowers. is this true or false

Chemistry

2 answers:

Afina-wow [57]3 years ago

6 0

The correct answer is true

Firdavs [7]3 years ago

4 0

The first pesticides relied on natural chemicals, such as those taken from chrysanthemum flowers is true

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The chemical bonds of carbohydrates and lipids have high potential energy because:

gladu [14]

Answer:

c. Many of their bonds are C-C and C-H

Explanation:

The majority of bonds in carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.

These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.

The correct answer is c.

4 0

3 years ago

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0

3 years ago

Identify all the examples that support the Law of Conservation of Mass through chemical changes.

wariber [46]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

N sub 2 +3H sub 2 rightwards arrow 2NH sub If 6 liters of hydrogen gas are used, how many liters of nitrogen gas will be needed

astraxan [27]

Answer:

2L of nitrogen gas will be needed

Explanation:

Based on the following reaction:

N₂ + 3H₂ → 2NH₃

<em>1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.</em>

<em />

If 6L of hydrogen (In a gas, the volume is directly proportional to the moles, Avogadro's law) react, the volume of nitrogen gas required will be:

6L H₂ * (1mol N₂ / 3 moles H₂) =

<h3>2L of nitrogen gas will be needed</h3>

7 0

3 years ago

How many grams of HNO3 are produced when 59.0 g of NO2 completely reacts?

Anettt [7]

Answer:

53.7 grams of HNO3 will be produced

Explanation:

Step 1: Data given

Mass of NO2 = 59.0 grams

Molar mass NO2 = 46.0 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 59.0 grams / 46.0 g/mol

Moles NO2 = 1.28 moles

Step 4: Calculate moles HNO3

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3

Step 7: Calculate mass HNO3

Mass HNO3 = 0.853 moles * 63.01 g/mol

Mass HNO3 = 53.7 grams

53.7 grams of HNO3 will be produced

3 0

3 years ago