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Alona [7]
3 years ago
11

1.) What are the two broad categories that elements can be divided into?

Chemistry
1 answer:
hoa [83]3 years ago
5 0

Answer:

1) Metals and nonmentals

2) Elements: Oxygen (O_{2} \\) , Nitrogen (N_{2}

3) Compounds: Carbon Dioxide CO_{2}, Methane CH_{4}, Nitrogen DioxideNO_{2}

Explanation:

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Okay So Blink your eyes. Was work done? Explain your answer.
Yuri [45]

Answer:

no work was needed. Blinking does not require using energy sometimes we blink without thinking.

Explanation:

7 0
3 years ago
Read 2 more answers
An ionic bond can be formed when one or more electrons are
xz_007 [3.2K]

Answer:

Are transferred completely from the valence shell of an element to the other

Explanation:

Basically, to form a chemical bond, you either transfer or you share. When you share, it is a case of covalent bonding which can be in several other forms. When there is a transfer, it is a case of ionic bonding.

The basic explanation for this is that while some atoms are electronically sufficient, some are electronically deficient. This means while some atoms are having an excess number of electrons, then some are having less number of electrons.

To satisfy both parties, there must be a transfer if electrons between the two parties. While the one with the excess numbers serves as the donor, the one with insufficient number of electrons serve as the acceptor

8 0
3 years ago
What are the evidences for suspecting the presence of waves?
Vsevolod [243]

Answer:

Well,

Conducting rods are good for detecting oscillating electric fields and conducting loops are good for detecting the presence of radio waves.  

5 0
3 years ago
On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of
Murrr4er [49]

Answer:

The empirical formula of the organic compound is  = C_2H_6S_1

Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

Moles of CO_2 gas at STP occupying 2.0 L = n

n\times 22.4L=2.0L

n=\frac{2.0 L}{22.4 L}=0.08929 mol

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of H_2O gas at STP occupying 3.0 L = n'

n'\times 22.4L=3.0L

n'=\frac{3.0 L}{22.4 L}=0.1339 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of SO_2 gas at STP occupying 1.0 L = n''

n''\times 22.4L=1.0L

n''=\frac{1.0 L}{22.4 L}=0.04464 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = \frac{0.08920 mol}{0.04464 mol}=2

Hydrogen =  \frac{0.2678 mol}{0.04464 mol}=6

Sulfur = \frac{0.04464 mol}{0.04464 mol}=1

The empirical formula of the organic compound is  = C_2H_6S_1

3 0
3 years ago
when 45 grams of copper (ii) carbonate are decomposed with heat how many grams of carbon dioxide will be produced
Maksim231197 [3]

Answer:

16.02 g

Explanation:

the balanced equation for the decomposition of CuCO₃ is as follows

CuCO₃ --> CuO + CO₂

molar ratio of CuCO₃ to CO₂ is 1:1

number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol

according to the molar ratio

1 mol of CuCO₃ decomposes to form 1 mol of CO₂

therefore 0.364 mol of  CuCO₃ decomposes to form 0.364 mol of CO₂

number of CO₂ moles produced - 0.364 mol

therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g

16.02 g of CO₂ produced

8 0
3 years ago
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