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VLD [36.1K]
3 years ago
9

Uestion 3 (1 point)

Chemistry
2 answers:
elena-s [515]3 years ago
6 0

Answer:

density=mass/volume.

density=14/18

=0.777777777 ~ 0.8g/ml.

It will float because the it lesser than the density of water.

Kaylis [27]3 years ago
5 0

formula for density of liquid = mass ÷ vol

= 14.0 g ÷ 18.0 cm³

= 0.77778 g/cm³ ( 5sf)

= 0.778 g/cm³ ( 3sf )

18.0 cm³ is the same as 18.0 ml

Therefore, the liquid will float on water as the water is denser than the liquid.

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ch4aika [34]
The correct answer is (1) one mole of NO2.

The gram formula mass is also known as the molar mass and is defined by the mass over one mole of a substance.

Hope this helps~
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3 years ago
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Why do gases diffuse more quickly than liquids?
Alborosie

Answer:

D)Gas particles move rapidly and have space between them.

Explanation:

Matter exists in three states namely: solids, liquids and gases. The particles contained in these three states are different from one another. In the gaseous state, the particles are FAR APART from one another i.e. space exists and they move at a very fast rate in contrast to the particles of a liquid, which have less space and move slower.

This rapid movement of gas particles within a less restricted space accounts for the reason why gaseous substances DIFFUSE more quickly than liquids.

5 0
3 years ago
112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.
dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

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

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0
2 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

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2 years ago
A selectively permeable membrane separates solution a and
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The water molecules will flow from b to a due to osmosis.

Osmosis is where water molecules will flow from a region of higher water potential to a region of lower water potential, through a selectively permeable membrane.

When the water molecule concentration is higher, it has a higher water potential top. Water potential is the tendency for them to flow to a lower region.

The net movement will stop until both sides of the solution has a same water potential.
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3 years ago
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