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Korvikt [17]
3 years ago
11

What is the difference between a do universal law and a scientific theory

Chemistry
1 answer:
Anika [276]3 years ago
4 0

A hypothesis is a limited explanation of a phenomenon; a scientific theory is an in-depth explanation of the observed phenomenon. A law is a statement based on an observed phenomenon or a unifying concept,


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Is BH2F a gas or a liquid and what is the boiling point
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How long the Earth takes to rotate on its axis and what Earth’s rotation causes?
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Manganese(III) oxide is a transition metal compound. The oxidation state of manganese in this compound is _______ , and the chem
Kisachek [45]

Answer:

Hello my Friend! The answer is: Manganese(III) oxide is a transition metal compound. The oxidation state of manganese in this compound is +3 , and the chemical formula of the compound is Mn2O3.

Explanation:

Manganese can have two oxidation states: +2 and +3, but in this case, the "(III)"  indicates that in this compound, the state of oxidation is +3.

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3 years ago
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Dustin is mixing concrete. A formula for concrete calls for 3/8 gal of water. Dustin wants to make 7/9 more concrete than the fo
slava [35]
We can use a ratio to solve this question. Lets refer to the amount formed by the formula as a serving
1 serving needs 3/8 gal
We want to create 7/9 servings extra, so
1 + 7/9 = 16/7
1 : 3/8
16/7 : x
1/(3/8) = (16/7) / x
8 / 3 = 16x / 7
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He needs to use 7/6 gallons of water.
5 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
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