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ddd [48]
3 years ago
5

What is the angle between the carbon-oxygen bond and one of the carbon-fluorine bonds in the carbonyl fluoride ( cof2 ) molecule

?

Chemistry
2 answers:
harina [27]3 years ago
7 0

Answer: More than 120^o.

Explanation:

In carbonyl fluoride molecule, carbon is sp^2 hybridized with planar geometry. Oxygen in doubled bonded to the carbon and two fluorine atoms are single bonded with carbon atom.

In sp^2 hybridization the angle between the two adjacent hybridized molecular orbitals is at 120^o.Since, lone pair present on the oxygen atoms will repel the carbon-fluorine bond pairs away by which angle between (O-C-F) will become more than 120^o.

kozerog [31]3 years ago
5 0
It has a trigonal planar structure. So, the calculated angle is 127.3 degrees.
You can see the structure in the picture.


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Scientific evidence is most likely to be consistent if it is based on data from
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Answer:

Random samples

Explanation:

It needs to be random so that there isn't bias that would skew the consistency

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Picking up a hot coal is an example of what type of heat transfer?
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Conduction should be the correct answer.
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A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to Fe²⁺ in acid and then titrating
neonofarm [45]

The mass percent of iron in the ore is 31.6%

<h3>Steps</h3>

8H⁺(aq) + 5Fe²⁺(aq) + MnO⁻₄(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(l)

V(MnO4^-)=39,32 mL

c(MnO4^-)=0,0319M

n(MnO4^-)=c*V=1,254308 mmol

n(Fe^{2+})=5*n(MnO4^-)=6,27 mmol

Equation

Fe+2H->Fe^+2 + H2

n(Fe)=n(Fe^{2+} )=6,27 mmol

m(Fe)=n*Ar=350,27mg=0,35027g

mass percent(Fe in ore)=m(Fe)/m(ore)*100

=31,61 percent

The mass percent of iron in the ore is 31.6%

<h3>What is the equation for the balance formula?</h3>

An equation for a chemical reaction is said to be balanced if both the reactants and the products have the same number of atoms and total charge for each component of the reaction. In other words, both sides of the reaction have an equal balance of mass and charge.

<h3>An illustration of a balanced equation</h3>

For instance, consider about the reaction: O2 (g) + 2Mg(s) = 2MgO (g) Two magnesium and two oxygen atoms are present in this reaction on both the reactant and product sides.

learn more about balanced equation here

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8 0
1 year ago
A student measures the mass of an 8 cm^3 block of brown sugar to be 12.9 g. What is the density of
Oduvanchick [21]

Answer:

1.6125 g/cm^3

Explanation:

Density= mass/volume

4 0
3 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
2 years ago
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