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Leya [2.2K]
3 years ago
15

How does a new hypothesis replace an order exclamation of something in the natural world

Chemistry
1 answer:
horrorfan [7]3 years ago
8 0
The knowledge gained by observing the natural world. Scientific ... New data can lead a scientist to change their hypothesis. A series of ... A. They always involves seeing something.
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What is the name of this hydrocarbon?
seraphim [82]
It's 2-methylpropane.
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4 0
4 years ago
How many grams are in 1.0 moles of baking soda?
kiruha [24]
The answer is 84.00661 grams
4 0
3 years ago
Give the characteristics of a strong acid.has a polar bondhas a weaker bond to hydrogenhas equilibrium far to the rightionizes c
Dominik [7]

Answer:

All of the above

Explanation:

<em>Give the characteristics of a strong acid.</em>

  • <em>Has a polar bond.</em> YES. A big difference in the electronegativity between the heteroatom and the hydrogen atom is associated with the strength of an acid.
  • <em>Has a weaker bond to hydrogen.</em> YES. A weaker bond to hydrogen makes it easier for it to break.
  • <em>Has equilibrium far to the right.</em> YES. The equilibrium of a strong acid is very shifted towards the products.
  • <em>Ionizes completely in aqueous solutions.</em> YES. A strong acid exists mostly in the ionic form in aqueous solution.
8 0
3 years ago
How many moles of O₂ are needed to burn 2.5 moles of CH₃OH?
IrinaK [193]

Answer:

3.75 moles

Explanation:

The chemical equation is 2CH₃OH + 3O₂ -> 2CO₂ + 4H₂O

2 moles of CH₃OH are burned by 3 moles of O₂

For 2.5 moles of CH₃OH are burned by x moles of O₂

Let's solve for x :

2*x=2.5*3 => 2*x=7.5 => x=3.75 moles of O₂ are needed to burn 2.5 moles of CH₃OH

5 0
3 years ago
A gas is contained in a thick-walled balloon. When the pressure changes from 39.1 bar to 87.0 bar the volume changes from blank
lidiya [134]

Answer: -

1.34 L

Explanation: -

Initial Pressure P 1 = 39.1 bar

Initial Temperature T 1 = 643 K

Let the initial volume be V 1.

Final pressure P 2 = 87.0 bar

Final temperature T 2 = 525 K.

Final volume V 2 = 0.492 L

Using the equation

\frac{P1 V 1}{T1} = \frac{P 2 V 2}{T2}

V 1 = \frac{P2 V2 T1}{P 1 T2}

Plugging in the values

We have

V 1 = 87 bar x 0.492 L x 643 K / (39.1 bar x 525 K)

= 1.34 L

Thus, a gas is contained in a thick-walled balloon. When the pressure changes from 39.1 bar to 87.0 bar the volume changes from 1.34 L to 0.492L and the temperature changes from 643K to 525K

4 0
4 years ago
Read 2 more answers
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