5.0g of aluminum fluoride decomposes into aluminum and oxygen. What mass of oxygen is formed?

Rutherford's research was primarily involved with: electron distribution defining the components of nuclei using electromagnetic

Nadya [2.5K]

Answer is: defining the components of nuclei.

Rutherford theorized that atoms have their charge concentrated in a very small nucleus.

This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).

Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.

That is different from Plum Pudding model, because it shows that most of the atom is empty space.

Rutherford conducted research that led to the first splitting of the atom in a nuclear reaction.

4 0

2 years ago

It is thought that Vitamin C might be an effective preventative

Anon25 [30]

Answer:

IF the fruit juice contains a high level of vitamin C, THEN the preventative effectiveness against common cold increases.

Explanation:

The hypothesis is a testable explanation of a scientific investigation. It aims at predicting the outcome of the experiment. One feature of the hypothesis is that it must be testable. The hypothesis is usually written in an "IF, THEN" format.

This question is regarding an experiment to test the amount of vitamin C in fruit juice. The vitamin C is thought to be an effective preventative against common cold. Hence, the hypothesis connects the effect on common cold (dependent variable) with the amount of vitamin C (independent variable). The hypothesis can be written as:

IF the fruit juice contains a high level of vitamin C, THEN the preventative effectiveness against common cold increases.

7 0

2 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

2 years ago

You have an aqueous solution of chromium(III) nitrate that you titrate with an aqueous solution of sodium hydroxide. After a cer

barxatty [35]

Answer:

The precipitate was chromium hydroxide, which then reacted with more hydroxide to produce a soluble complex, Cr(OH)4

Explanation:

The following reaction takes place when chromium(III) nitrate reacts with NaOH:

$Cr(NO)_{3}$ +3 NaOH → $Cr(OH)_{3}$ (s)+ $NaNO_{3}$

The precipitate that is formed is chromium hydroxide, $Cr(OH)_{3}$

When more NaOH is added, the precipitate reacts with it which then results in the formation of a soluble complex ion:

$Cr(OH)_{3}$ (s) + $OH^{-}$ (aq) → $Cr(OH)_{4} ^{-}$ (aq)

$Cr(OH)_{4} ^{-}$ is soluble complex ion

7 0

3 years ago

Chemistry

postnew [5]

There are 48.72 g Fluorine ions

<h3>Further explanation </h3>

Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements

In the same compound, although from different sources and formed by different processes, it will still have the same composition/comparison

%F in CaF₂ :

$\tt \%F=\dfrac{2.Ar~F}{MW~CaF_2}\times 100\%\\\\\%F=\dfrac{2.19}{78}\times 100\5=48.72\%$

mass of Fluorine :

$\tt 48.72\%\times 200=97.44~g$

So mass Fluorine ions(2 ions F in CaF₂⇒Ca²⁺+2F⁻) :

$\tt =\dfrac{97.44}{2}=48.72~g$

7 0

2 years ago