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juin [17]
3 years ago
13

How does the concentration (amount) of carbon dioxide in the affect climate?

Chemistry
1 answer:
Fiesta28 [93]3 years ago
7 0

Answer:

A

Explanation:

When the carbon dioxide concentration goes down, temperature goes down. A small part of the correspondence is due to the relationship between temperature and the solubility of carbon dioxide in the surface ocean, but the majority of the correspondence is consistent with a feedback between carbon dioxide and climate.

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Uranium, an important component of both nuclear weapons and nuclearreactors, has two major isotopes, U-238, which has a half-lif
elixir [45]

Answer:

Hello your question is poorly written below is the well written question

Uranium, an important component of both nuclear weapons and nuclear reactors, has two major isotopes, U-238, which has a half-life of  approximately 4.5 billion years, and U-235, which has a half-life of approximately 700 million years. Both were present in equal amounts at the time of the creation of the Earth, 4.5  billion years ago. How many years after the creation of the Earth had the amount of radiation from uranium decayed to half the amount present at the time of the creation of the Earth

Answer : 140 billion years

Explanation:

Given that :

U-238   h1/2 = 4.5 billion years

U-235   h1/2 = 700 million years

At the beginning both Isotopes where present in equal amount

Determine the T years before the amount of Uranium decays to Half

T =   ?    N'2 = N1 / 2

we know that N = No ( 1/2 )^h   where h = time / half-life time

attached below is the detailed solution of the given problem

6 0
2 years ago
Kathleen makes iced coffee by adding ice to her room temperature coffee in a glass. What statement explains what is taking place
Ganezh [65]

Answer:

The answer is the heat is transferred from the ice to the coffee.

Explanation:

Hot molecules tend to transfer to colder molcules, which causes the ice to melt and makes the coffee colder.

Hope this helps! :)

3 0
1 year ago
Compare the root mean square speeds of O2 and UF6 at 65 degree Celsius
gayaneshka [121]
<h3>Answer:</h3>

The root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.

<h3>Solution and Explanation:</h3>
  • To find how fast molecules or particles of gases move at a particular temperature, the root  mean square speed is calculated.
  • Root mean square speed of a gas is calculated by using the formula;

       Root mean square=\sqrt{ \frac{3RT}{M}

       Where R is the molar gas constant, T is the temperature and M is the molar mass of gas in Kg.

<h3>Step 1: Root mean square speed from O₂</h3>

Molar mass of Oxygen is 32.0 g/mol or 0.032 kg/mol

Temperature = 65 degrees Celsius or 338 K

Molar gas constant = 8.3145 J/k.mol

Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.032}

                                        = 513.289 m/s

<h3>Step 2: Root mean square speed of UF₆   </h3>

The molar mass of UF₆ is 352 g/mol or 0.352 kg/mol        

Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.352}

                                                 = 154.762m/s

Therefore; the root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.

7 0
2 years ago
The half-life period for a first order reaction is independent of
maria [59]
I believe the answer is concentration

Sorry if I’m wrong hope this helps
4 0
2 years ago
Read 2 more answers
). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50
Nikitich [7]

Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C =  2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568  g C

Mass of H = 0.6955 g H₂O  × (2.016 H/18.02 g H₂O)  = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

=====

2. Convert these masses to moles.

Moles  C = 0.5568  × 1/12.01  = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654   × 1/16.00 = 0.054 09

=====

3. Find the molar ratios.

Moles  C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36   = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

======

4. Multiply the ratios by a number to make them close to integers

C  = 1        × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6  = 7.001

=====

5. Round the ratios to integers

C:H:O =6:10:7

=====

6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

=======

7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

=====

8. Divide the molecular mass by the empirical formula mass.  

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

=====

9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

7 0
3 years ago
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