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IrinaK [193]
3 years ago
10

Scientist can use trees to look at climates of the past. How? What information can they gather and how do they gather it? Explai

n.
Chemistry
1 answer:
andrezito [222]3 years ago
6 0

Explanation:

Scientist use trees a whole lot to look at climate of the past by examining tree rings.

These are layers of cambium in each successive years formed. They have an annual growth pattern and are known as tree rings.

Tree rings can be used to decipher the age of a tree.

  • These three rings can be used to interpret climatic patterns.
  • During a wet climate, the tree rings are more robust and bigger.
  • In a dry climate, the rings are thinner.
  • These alternating patterns can be used to decipher the climatic signatures in a tree.
  • Sometimes, it is possible to evaluate some certain isotopes that are useful in climatic studies.

learn more:

Climate change brainly.com/question/7824762

#learnwithBrainly

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Answer:

Cytoplasm.

Explanation:

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What chemical is added to natural gas in case of gas leaks?
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A sample of sugar contains 1.505 × 1023 molecules of sugar. How many moles of sugar are present in the sample?
alisha [4.7K]
1 mole ------------ 6.02 x10²³ molecules
? mole ----------- 1.505 x10²³ molecules

1.505x10²³ / 6.02x10²³ => 0.25 moles

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4 0
3 years ago
The first-order decomposition of cyclopropane has a rate constant of 6.7 x 10-4 s-1. if the initial concentration of cyclopropan
vaieri [72.5K]
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:

ln [A] = -kt + ln [A]_o, where,

k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species

For this problem, we simply substitute the known values to the equation as in:

ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M) 

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5 0
3 years ago
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Help Pls!
Salsk061 [2.6K]

Answer:

d= 50.23 g/cm³

Explanation:

Given data:

radius = 137.9 pm

mass is = 5.5 × 10−22 g

density = ?

Solution:

volume of sphere= 4/3π r³

First of all we calculate the volume:

v= 4/3π r3

v= 1.33× 3.14× (137.9)³

v= 1.33 × 3.14 × 2622362.939 pm³

v= 1.095 × 10∧7 pm³

v= 1.095 × 10∧-23 cm³

Formula:

Density:

d=m/v

d= 5.5 × 10−22 g/ 1.095 × 10∧-23 cm³

d= 5.023 × 10∧+1 g/cm³

d= 50.23 g/cm³

8 0
3 years ago
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