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djverab [1.8K]
3 years ago
14

A scientist performed an experiment with a mouse where the mouse was moving along a track with a new experimental unscented chee

se at the end of the track. On the graph the end of the track was 8 meters and the mouse started at 3 m along the track. After 5 seconds what was the velocity of the mouse?
A scientist performed an experiment with a mouse where the mouse was moving along a track with a new experimental unscented cheese at the end of the track. On the graph the end of the track was 8 meters and the mouse started at 3 m along the track. After 5 seconds what is the distance the mouse traveled?

Chemistry
1 answer:
kow [346]3 years ago
7 0

Answer:

Explanation:

a achually im not sure but i need points plz dont report me, bc no one els is even answering sooo plz there is no reason to flag, i will give u memes

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Answer:

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What transition energy corresponds to an absorption line at 527 nm?​
Ede4ka [16]

Answer:

E = 3.77×10⁻¹⁹ J

Explanation:

Given data:

Wavelength of absorption line = 527 nm (527×10⁻⁹m)

Energy of absorption line = ?

Solution:

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E = hc/λ

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c = speed of wave = 3×10⁸ m/s

by putting values,

E = 6.63×10⁻³⁴ Js ×  3×10⁸ m/s / 527×10⁻⁹m

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3 years ago
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An analysis of an ionic compound is found to contain 92 grams of sodium (Na) and 32 grams of oxygen (O). Determine the empirical
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Acetic acid, CH3COOH, can be produced by bubbling oxygen gas into acetaldehyde, CH3CHO, in the presence of
slamgirl [31]

Explanation:

The balanced equation for the reaction is given as;

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If 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel, (a)

how many grams of acetic acid will be produced?

First thing's first, we have to find he limiting reactant. This is done by comparing the number of moles of the reactants.

From the equation of the reaction;

2 mol of CH3CHO reacts with 1 mol of O2

From the masses given;

Number of moles = Mass / Molar mass

CH3CHO;

Number of moles = 20 / 44.0526 = 0.454 mol

O2;

Number of moles = 10 / 32 = 0.3125 mol

The limiting reactant is CH3CHO because O2 would be in excess.

Back to the question;

2 mol of CH3CHO produces 2 mol of CH3COOH  

0.454 mol would produce x

Solving for x;

x = 0.454 * 2 / 2 = 0.454 mol

Converting to mass;

Mass = number of moles* Molar mass

Mass = 0.454 mol *  60.052 g/mol = 27.26 grams

(b) how many grams of the excess reactant remain after the reaction is

complete

The excess reactant is O2

Number of moles left = Initial Number of moles - Number of moles that reacted

Number of moles left =  0.3125 mol - (0.454 mol / 2)

Number of moles left = 0.0855 mol

Converting to mass;

Mass = 0.0855 mol * 32 g/mol = 2.736 grams

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