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zvonat [6]
3 years ago
5

Periods occurred during the Mesozoic Era? ​

Chemistry
2 answers:
svetoff [14.1K]3 years ago
8 0

Triassic
Jurassic
Certacous

The Mesozoic is divided into three time periods: the Triassic (245-208 Million Years Ago), the Jurassic (208-146 Million Years Ago), and the Cretaceous (146-65 Million Years Ago).
Sergeeva-Olga [200]3 years ago
6 0
Triassic Jurassic and Cretaceous
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One important sugar that results from photosynthesis is cellulose. True or false
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It true because in photosynthesis, you need celluos
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Where is the energy for cell division generated?
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If a neutral atom with 88 electrons undergoes 2 rounds of beta minus decay, what will be the new element?
Charra [1.4K]

Answer:

The new element will be thorium-226 (²²⁶Th).

Explanation:

The beta decay is given by:

^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + \beta^{-} + \bar{\nu_{e}}

Where:

A: is the mass number

Z: is the number of protons  

β⁻: is a beta particle = electron

\bar{\nu_{e}}: is an antineutrino

The neutral atom has 88 electrons, so:

e^{-} = 88 = Z

Hence the element is radium (Ra), it has A = 226.

If Ra undergoes 2 rounds of beta minus decay, we have:

^{226}_{88}Ra \rightarrow ^{226}_{89}Ac + \beta^{-} + \bar{\nu_{e}}    

^{226}_{89}Ac \rightarrow ^{226}_{90}Th + \beta^{-} + \bar{\nu_{e}}

Therefore, if a neutral atom with 88 electrons undergoes 2 rounds of beta minus decay the new element will be thorium-226 (²²⁶Th).

I hope it helps you!    

8 0
3 years ago
If you heat your melting point tube too quickly, what will your observed melting point be in comparison to the actual melting po
solniwko [45]

Explanation:

When conducting a melting point experiment, if we were to heat a sample quickly. Large amount heat is provided instantly which would melt the crystals in the tube very quickly, even before the temperature of the thermometer reaches to that level. So the observes melting point would be much lower than the actual melting point when sample is heated slowly.

5 0
3 years ago
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. 2A(g)
monitta

Answer:

The value of the missing equilibrium constant ( of the first equation) is 1.72

Explanation:

First equation: 2A + B ↔ A2B   Kc = TO BE DETERMINED

 ⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]

Second equation: A2B + B ↔ A2B2   Kc= 16.4

⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]

Third equation:  2A + 2B ↔ A2B2     Kc = 28.2

⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²

If we add the first to the second equation

2A + B + B ↔ A2B2   the equilibrium constant Kc will be X(16.4)

But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2)   with Kc = 28.2

So this means: 28.2 = X(16.4)

or X = 28.2/16.4

X = 1.72

with X = Kc of the first equation

The value of the missing equilibrium constant ( of the first equation) is 1.72

7 0
4 years ago
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