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seraphim [82]
3 years ago
10

What type of energy warms earths surface?

Chemistry
1 answer:
Scrat [10]3 years ago
8 0
A. solar energy

the earth is warmed by the sun, which is solar (or infrared).
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A certain reaction has an activation energy of 67.0 kJ/mol and a frequency factor of A1 = 4.20×1012. What is the rate constant,
MrRa [10]

Answer:

  • K = 7.00

Explanation:

<u>1) Arrhenius equation</u>

Arrhenius equation shows the relation between activation energy, temperature, and the equilibrium constant.

This is the equation:

       K=Ae^{-Ea/RT}

Where:

  • K is the equilibrium constant,
  • A is the frequency factor,
  • Ea is the activation energy (in J/mol),
  • T is the temperature in kelvins (K), and
  • R is the universal constant.

<u></u>

<u>2) Substitute, using the right units, and compute:</u>

  • A = 4.20 × 10¹² (dimensionless)
  • Ea = 67.0 kJ/mol = 67,000 J/mol
  • T = 24.0°C = 24.0 + 273.15 K = 297.15 K
  • R = 8.314 J/K mol

K=Ae^{-Ea/RT}=(4.20)(10)^{12}e^{-67,000J/mol/(8.314J/molK.297.15K)}

  • K = 7.00
5 0
3 years ago
The conjugate acid of ch3nh2 is ________. the conjugate acid of ch3nh2 is ________. ch3nh2+ ch3nh2 ch3nh3+ ch3nh+ none of the ab
Verizon [17]
The conjugate acid of ch3nh2 is ch3nh3+<span>.
</span>For example methylamine in water chemical reaction: 
CH₃NH₂(aq)+ H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq).
According to  Bronsted-Lowry theory acid are donor of protons and bases are acceptors of protons (the hydrogen cation or H⁺). Methylamine (CH₃NH₂) is Bronsted base and it can accept proton and become conjugate acid (CH₃NH₃⁺).
3 0
3 years ago
A wooden boat is discovered just south of the Graet Pyramid in Egypt had a Carbon-14 content of approximately 25% of that found
wolverine [178]

Because 25% is half of 0.5 and the half life of {}^{14}\text{C} is given as 5,715, the time it takes to reach this amount should just be double the half life: 11430 years.

However, in other cases the problem may not have easy numbers, so below is a more mathematical approach.

The half life formula is A=A_0*(\frac{1}{2} )^{\frac{t}{h} }.

Solving for t gives:

t=\frac{h*\ln(\frac{A}{A_0} )}{\ln(\frac{1}{2} )}

Given information:

A=0.25A_0\\h=5715\text{ yrs}

Plug in given values and solve for t.

t=\frac{5715 *\ln(0.25) }{\ln(\frac{1}{2} )} =11430 \text{ yrs}

With correct significant figures, this would be 11000 years.

6 0
3 years ago
4. Which statement is true about alkali metals?
Svet_ta [14]
I think its most likely to be A,
8 0
4 years ago
Read 2 more answers
Predict what would happen to the volume of the gas if the pressure in Step 3 was increased to 800 kPa.
Inga [223]
The volume you would expect the gas to occupy if the pressure is increased to 40
The kpa would be 50 liters
8 0
3 years ago
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