You collect 14.2 liters of CO gas from an experiment. How many molecules would be in the sample?

What characteristics or characteristics contribute to the attraction between earth and the moon

OLEGan [10]

Answer:

The moon's gravity pulls at the Earth, causing predictable rises and falls in sea levels known as tides. To a much smaller extent, tides also occur in lakes, the atmosphere, and within Earth's crust. High tides are when water bulges upward, and low tides are when water drops down.

5 0

3 years ago

Read 2 more answers

I know how to solve it with D=M/V and M1V1 however the answer isn’t correct. Help me please

lara31 [8.8K]

Answer:

23.28 g of O2.

Explanation:

We'll begin by calculating the mass of hexane. This can obtain as follow:

Volume of hexane = 10 mL

Density of hexane = 0.66 g/mL

Mass of hexane =?

Density = mass /volume

0.66 = mass of hexane /10

Cross multiply

Mass of hexane = 0.66 x 10

Mass of hexane = 6.6 g

Next, we shall write the balanced equation for the reaction. This is given below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:

Molar mass of C6H14 = (12.01x6) + (1.008 x 14)

= 72.06 + 14.112

= 86.172 g/mol

Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608 g

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.

Therefore, 23.28 g of O2 is needed for the reaction.

8 0

3 years ago

An alkane with the formula C6H14 can be prepared by hydrogenation of either of two precursor alkenes having the formula C6H12.

ankoles [38]

Two precursor alkenes

H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene

H₃C   CH₃
I   I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene

alkane

H₃C CH₃
  I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane

H₃C  CH₃ H₃C CH₃
I   I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃

H₃C CH₃ H₃C   CH₃
I    I     I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃

6 0

3 years ago

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How to solve 5gNaCl= MolE NaCl

valentinak56 [21]

0.086mole

Explanation:

The problem is very simple. It involves conversion from mass to number of moles of NaCl

What is mole?

A mole is a unit of measuring chemical substances. It is the amount that contains the avogadro's number of particles.

How is it related and can be converted from mass;

Number of moles of a substance = $\frac{mass}{molar mass}$

Since we know the mass, we can find the molar mass and put into this equation;

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Number of moles of NaCl = $\frac{5}{58.5} =$ 0.086mole

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

4 0

3 years ago

Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.

N76 [4]

Answer:

1 . $Al_2O_3$

2. $TiO_2$

Explanation:

The more stable the ionic compound, the more is it lattice energy.

The more the charge on the cation and the anion, the greater is the lattice energy.
The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide ( $Sc_2O_3$ ) is an oxide in which $Sc^{3+}$ behaves as cation and $O^{2-}$ behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . $Al_2O_3$

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation $Al^{3+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

2. $TiO_2$

This is because the charge on the cation $Ti^{4+}$ is greater than that of $Sc^{3+}$ and also the size of the cation $Ti^{4+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

3 0

3 years ago