It is possible for gases in the atmosphere to change the behavior of energy on earth. Agree or disagree

Answers

Answer 1
Answer:

The green house gases in the atmosphere can change the behavior of energy on earth. Thus i agree with this statement.

What is green house effect?

The green house effect is defined as a process by which the radiations from the sun are absorbed by the green house gases like methane, Chloro fluoro carbons and not reflected back into the space. This makes the surface as insulator and prevents it from freezing.

Due to the increased level of green house gases, the temperature of the earth increases considerably and thereby causes global warming, depletion of ozone layer, smog and air pollution, etc.

It is the natural green house effect which affect the behaviour of heat energy radiated by the sun. The green house gases never let the radiations escape from the earth and increase the surface temperature of earth. This leads to global warming.

Thus green house gases affects the solar energy.

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Answer 2
Answer:

Answer:

I agree

Explanation:

because the sunlight passes through the atmosphere and is absorbed by earth's surface.


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The figure below represents a reaction.What type of reaction is shown?SO3+ H2SO4 →→ HSO4
synthesis
decomposition
single displacement
double displacement

Answers

SO₃+ H₂SO₄ → H₂SO₄ This reaction is synthesis type of reaction, because that would be combination (synthesis) A + B → AB. Therefore, option A is correct.

What are the types of reaction ?

There are five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion.

Biochemical reactions are chemical reactions that occur within living things. Metabolism refers to the sum of all biochemical reactions in an organism. Exothermic and endothermic chemical reactions are both part of metabolism.

Hydration is the process of combining water molecules with another substance to form a single, new compound. SO3 is an acidic oxide that reacts with and dissolves in water to form sulfuric acid, H2SO4.

Thus, option A is correct.

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Are you sure it isn’t SO3+H2O = H2SO4 because that would be combination (synthesis) A+ B=AB

Or SO3 + H2SO4 = H2S2O7
Because that would also be synthesis

How many joules are released when one atom of americium 241, the isotope used in ionization-type smoke detectors, undergoes apha emission?

Answers

When Americium (Am-241) undergoes alpha decay(He-4) it forms neptunium (Np-237) based on the following pathway:

²⁴¹Am₉₅ → ²³⁷Np₉₃ + ⁴He₂

The energy released in given as:

ΔE = Δmc²

where Δm = mass of products - mass of reactants

                 = [m(Np-237) + m(He-4)] - [m(Am-241)]

                 = 237.0482+4.0015-241.0568 = -0.0073 g/mol = -7.3 * 10⁻⁶ kg/mol

ΔE = -7.3*10⁻⁶ kg/mol * (3*10⁸ m/s)² = -5.84*10¹¹ J/mol



Based on your Lewis structures for the postlab assignment, which molecules below have a three dimensional structure? (Select all that apply.)(A) CO2
(B) H2S
(C) SO3
(D) PCl3
(E) CH2Cl2
(F) NO2

Answers

Answer:

D and E

Explanation:

D. PCl3

E. CH2Cl2

Why can't 1−methylcyclohexanol be prepared from a carbonyl compound by reduction? select the single best answer?

Answers

1−methylcyclohexanol is a tertiary alcohol. Tertiary Alcohols are synthesized by either reacting Ketone with Organometallic compounds like Grignard reagent or by hydration of substituted alkenes. 1−methylcyclohexanol can not be synthesized by reduction of carbonyl compound because it is not possible to have a starting carbonyl compound having carbonyl group along with three other alkyl groups (as carbon can only form 4 bonds).

Result:
           Tertiary alcohols don't contain a hydrogen atom at carbon attached to hydroxyl group that is why it is not possible to synthesize 
1−methylcyclohexanol by reduction of carbonyl compound.

By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution. A stock solution of potassium permanganate (KMnO4) was prepared by dissolving 13.0g KMnO4 with DI H2O in a 100.00-mL volumetric flask and diluting to the calibration mark. Determine the molarity of the solution Molarity= O.822 M

Answers

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=(0.823 M* 11.00 mL)/(50 ,00 mL)=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}* Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=(13.0 g)/(158 g/mol* 0.100 L)=0.823 mol/L

0.823 Molar is the molarity of the solution.

Final answer:

To determine the molarity of the resulting solution, we can use the formula M1V1 = M2V2. Plugging in the given values, we find that the molarity of the resulting solution is 0.180 MM.

Explanation:

To determine the molarity of the resulting solution, we need to use the formula:

M1V1 = M2V2

Where M1 is the molarity of the stock solution, V1 is the volume of the stock solution used, M2 is the molarity of the resulting solution, and V2 is the final volume of the resulting solution.

Using the given values, we have:

M1 = 0.823 MM

V1 = 11.00 mL

V2 = 50.00 mL

Substituting these values into the formula, we can find the molarity of the resulting solution.

M2 = (M1 * V1) / V2

Plugging in the values:

M2 = (0.823 MM * 11.00 mL) / 50.00 mL = 0.180 MM

The molarity of the resulting solution is 0.180 MM.

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How many grams of sucrose must be added to 375 mL of watertoprepare a 2.75m/m percent solution of sucrose?

Answers

Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.

Explanation :

As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.

Mass of solution = 100 g

Mass of sucrose = 2.75 g

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 100 g - 2.75 g

Mass of water = 97.25 g

First we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}* \text{Volume of water}

Density of water = 1.00 g/mL

Volume of water = 375 mL

\text{Mass of water}=1.00g/mL* 375mL=375g

Now we have to calculate the mass of sucrose in 375 g of water.

As, 97.25 grams of water contain 2.75 grams of sucrose

So, 375 grams of water contain (375)/(97.25)* 2.75=10.6 grams of sucrose

Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.

To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.

To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.

Here's how you can calculate it:

1. Convert the volume of water to grams, considering the density of water:

  Density of water ≈ 1 g/mL

  Mass of water = Volume of water × Density of water

  Mass of water = 375 mL × 1 g/mL = 375 g

2. Determine the desired mass of sucrose as a percentage of the total mass:

  Desired m/m percent = 2.75%

3. Calculate the mass of sucrose needed:

  Mass of sucrose = (Desired m/m percent / 100) × Total mass

  Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)

4. Rearrange the equation to solve for the mass of sucrose:

  Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))

Now, calculate:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - 0.0275)

Mass of sucrose ≈ (2.75 / 100) × (375 g) / 0.9725

Mass of sucrose ≈ (2.75 × 375 g) / 100 / 0.9725

Mass of sucrose ≈ (1031.25 g) / 0.9725

Mass of sucrose ≈ 1061.98 g

So, approximately 1062 grams of sucrose must be added to 375 mL of water to prepare a 2.75 m/m percent solution of sucrose.

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