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3 years ago

If oxygen, symbol o, undergoes a covalent bond it wants to form how many bonds

Chemistry

2 answers:

Marysya12 [62]3 years ago

7 0

Answer: 2.

Explanation:

1) The number of bonds that an atom is willing to form may be predicted in terms of the octet rule.

2) Atomic number of oxygen is 8, which implies that the atom has 8 electrons.

3) The electron configuration of oxygen is 1s² 2s² 2p⁴.

4) So, oxygen has 6 electrons in its valence shell.

5) The octet rule states that any atom will try to reach the configuration of the closest noble gas by "gaiing" or "giving" the electrons to have the 8 electrons in the outer shell.

6) Then oxygen will be willing to share two electrons with other atoms to complete, along with its own 6 electrons, the 8 electrons, which leads to the configuration of the noble gas Ne.

7) Each electron shared contributes to the formation of a bond. That is why by sharing two electrons, oxygen will form two covalent bonds.

polet [3.4K]3 years ago

5 0

Oxygen has 16 bonds in total

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1) How many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7g Fe? Show your work.

klasskru [66]

1.) is157.7 g
moles Fe = 209.7 g/ 55.847 g/mol=3.75
the ratio between Fe and CO is 2 : 3
moles CO = 3.75 x 3 /2 =5.63
mass CO = 5.63 mol x 28.01 g/mol=157.7 g

2.) is 1.06 moles
48.7/23 = 2.12 moles sodium
2.12/2 x 24 = 25.44dm^3 hydrogen = 1.06 moles 
1.06 X 6.02x10^23 = 1.204x10^24 molecules of hydrogen.

3.) is 91.8
8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced. 8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically. % yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8

6 0

3 years ago

Read 2 more answers

A balloon with a volume of 2.0 L is filled with a gas at 3 atmospheres. If the pressure is reduced to 0.5 atmospheres without a

creativ13 [48]

Answer:

V2 = 12 L

Explanation:

Given the following data;

Initial volume = 2L

Initial pressure = 3 atm

Final pressure = 0.5 atm

To find the new volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Substituting into the equation, we have;

$3 * 2 = 0.5V_{2}$

$6 = 0.5V_{2}$

$V_{2} = \frac {6}{0.5}$

$V_{2} = 12$

V2 = 12L

Therefore, the new volume is 12 Liters.

6 0

2 years ago

In the lab you weigh out 76.02 g of Iron (Fe). How many moles of Iron do you have in the sample. (Your answer must have a unit..

son4ous [18]

Answer:

We have 1.361 moles in the sample

Explanation:

Mass of iron = 76.02g

Molar mass of iron = 55.845 g/ mole ( This we can find in the periodic table, and menas that 1 mole of iron has a mass of 55.845 g).

To calculate the number of moles we will use following formula:

moles (n) = mass / molar mass

moles iron = 76.02g / 55.845 g/ mole

moles iron = 1.36127 moles

To use the correct number of significant digits we use the following rule for multiplication and division :

⇒ the number with the least number of significant figures decides the number of significant digits.

⇒76.02 has 4 digits ( 2 after the comma) and 55.845 has 5 digits (3 after the comma).

⇒ this means 1.361 moles

We have 1.361 moles in the sample

3 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

What is the proper solution to the following equation? Answer with correct significant figures. (23.1 + 5.61 + 1.008) × 7.6134 ×

OleMash [197]

(23.1 + 5.61 + 1.008) × 7.6134 × 8.431

= 29.718 × 7.6134 × 8.431

=1907.55608

The equation is solved in the manner that the term in the bracket is added first then mutiplication is done as according to Bodmas rule, if an equation contains brackets that need to be solved first then other operation that is division,multiplication, addition and subtraction are performed from left to right .

5 0

3 years ago