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monitta
3 years ago
14

What is the Bo element plz help

Chemistry
2 answers:
Effectus [21]3 years ago
8 0
There is no Bo element, however there is a B element which is Boron

marishachu [46]3 years ago
5 0

Hi there!

Sorry to inform you, there is NO Bo element.

However, there is Boron (B),

Beryllium (BE),

Bromine (BR),

Bismuth (BI),

and Bohrium (Bh)

These are only a few elements.

Hope this helps! :)

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Someone please help! I will give Brainliest to who answers the best! Write an explanation of how mass determines (or affects) th
xxTIMURxx [149]
Gravity is the force of motion pulling down objects to the ground. If there was no gravity, everyone would walking as if they were on the moon.


Mass is what gravity needs. If an object has a little amount of mass, gravity will be able to easily bring it to the ground.


If an object has a very huge amount of mass, gravity will still be able to bring it to the ground but it will be hard.


For example: An airplane has a HUGE amount of mass. Gravity pulls it down but the airplane needs to be steering up in order for it to be straight. Gravity is applied on the airplane when it is landing.

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If an object has little mass and a table is in the way of gravity pulling it down to the ground, the object will stay on the table. Like a plate of food on a table.

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Glad to help! :) :D

8 0
3 years ago
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

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3 years ago
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