What is the generic equation for a combustion reaction?

What do you think the axis of rotation means?

larisa86 [58]

Answer:

it means the earth is rotating-

Explanation:

5 0

2 years ago

If the plum pudding model of the atom was correct, what should the results of Rutherford’s experiment be?

shusha [124]

Answer:most of the positively charge particles should be bounce back at a range of angles as they collide with the atoms in the foil; only a few should pass straight through the foil

Explanation:

6 0

3 years ago

The decay of which radioisotope can be used to estimate the age of the fossilized remains of an insect?

Dmitriy789 [7]

(4) C-14 Carbon is found in all living organisms.

3 0

3 years ago

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Is O2 always a double bond?

Dima020 [189]

O-O single bonds and H-O single bonds in $\text{H}_2\text{O}_2$ molecules.
H-O single bonds in $\text{H}_2\text{O}$ molecules.
O=O double bonds in $\text{O}_2$ molecules.

<h3>Explanation</h3>

How many valence electrons do atoms in each molecule need for them to be stable?

Each H atom needs two valence electrons to be stable.
Each atom of an element other than H needs eight valence electron to be stable.

There are two H atoms and two O atoms in an $\text{H}_2\text{O}_2$ molecule. Atoms in each $\text{H}_2\text{O}_2$ need $2 \times 1 + 2\times 2 = 6$ more electrons to be stable.
There are two H atoms and one O atom in an $\text{H}_2\text{O}$ molecule. Atoms in each $\text{H}_2\text{O}$ molecule need $2 \times 1 + 2= 4$ more electrons to be stable.
There are two O atoms in an $\text{O}_2$ molecule. Atoms in each $\text{O}_2$ molecule need $2 \times 2 = 4$ more electrons to be stable.

How many chemical bonds in each molecule?

Each chemical bond adds one valence electron to each bonding atom. Each chemical bond connects two atoms. As a result, each chemical bond adds two valence electrons to the molecule.

Each $\text{H}_2\text{O}_2$ molecule needs $6 / 2 = 3$ chemical bonds.
Each $\text{H}_2\text{O}$ molecule needs $4 / 2 = 2$ chemical bonds.
Each $\text{O}_2$ molecule needs $4 /2= 2$ chemical bonds.

What chemical bonds are these? Again, each H atom needs only one more valence electron to be stable. It will share only one electron with O and form one H-O bond. The rest of the chemical bonds are between O atoms.

There are two H atoms in each $\text{H}_2\text{O}_2$ molecule, which form two H-O bonds. Two of the three chemical bonds in this molecule are H-O. The other is an O-O single bond between the two O atoms.
There are two H atoms in each $\text{H}_2\text{O}$ molecule, which form two H-O bonds. Both chemical bonds in this molecule are H-O. There's no O-O bond in this molecule.
There is no H atom in $\text{O}_2$ molecules. Both chemical bonds are between O atoms. However, there are only two O atoms. There must be two chemical bonds between the two O atoms. That bond will be an O-O double bond.

3 0

3 years ago

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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago