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abruzzese [7]
3 years ago
7

What were schools like the first half of Henry's life? How do you think this affect his education? Answer BOTH questions in comp

lete sentences.
Chemistry
1 answer:
Lorico [155]3 years ago
3 0

Answer:

Not sure.

Explanation:

Who is Henry?

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What is the smallest part of an element?
GenaCL600 [577]

An atom is the smallest unit of matter that has the properties of an element. It is composed of a dense core called the nucleus and a series of outer shells occupied by orbiting electrons. The nucleus, composed of protons and neutrons, is at the center of an atom.

Explanation:

8 0
3 years ago
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What is the purpose of graphics in scientific articles
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Answer:

Graphics can sometimes convey more information in a brief amount of space than an author can explain in a paragraph.

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3 years ago
Which particle has a mass of approximately 1 atomic mass unit?
labwork [276]
The answer is D) a neutron.

When we say an atom's mass is, like 5 atomic mass units actually we are saying that the total number of the neutrons and protons in its nucleus is 5.

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Mass of a neutron is approximately 1 atomic mass unit.
7 0
3 years ago
A 100g sample of Carbon-14 has a half-life of 5 years. How much Carbon-14 is left after 10 years?
attashe74 [19]

Answer: The amount of carbon-14 left after 10 years is 25 g

Explanation:

Formula used :

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = ?

a_o = Initial amount of the reactant = 100 g

n = number of half lives =\frac{\text {given time}}{\text {half life}}=\frac{10}{5}=2

Putting values in above equation, we get:

a=\frac{100g}{2^2}

a=\frac{100g}{4}=25g

Therefore, the amount of carbon-14 left after 10 years is 25 g

6 0
3 years ago
For the reaction CO+2H2=CH3OH at 700 K, equilibrium concentrations are [H2]=0.072 M, [CO]= 0.020M, and [CH3OH]= 0.030 M. Calcula
bekas [8.4K]

The balanced equation for the reaction is

CO(g) + 2H₂(g) ⇄ CH₃OH(g)

 

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

   k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²

</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M

 

From substitution,

   k = 0.030 M / 0.020 M x (0.072 M)²

   k = 289.35 M⁻²

<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².

<span> </span>

3 0
3 years ago
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