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3 years ago

Predict what you need to do do to use boyle’s law on a macroscopic scale

Chemistry

1 answer:

Verizon [17]3 years ago

4 0

Answer:

Explanation:

Boyle's law states that:

"For an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume".

Mathematically:

$pV=const.$

where

p is the pressure of the gas

V is the volume of the gas

The important condition in order to verify Boyle's law in an experiment is that the temperature of the gas must remain constant. This means that the temperature of the gas during the experiment must be kept constant. This can be achieved, for example, by placing the apparatus containing your gas (for example, a glass tube) inside a thermal bath: for instance, a large amount of water kept at constant temperature. This way, the gas inside the tube will also have the same temperature of the water, which is constant.

Learn more about ideal gases:

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A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called

grandymaker [24]

Answer:
A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called cis Isomer.

Explanation:
Geometric isomerism takes place about the double bond in alkenes when the alkyl groups are either situated at the same side (cis) or are situated opposite (trans) to each other.

Example:
cis-2-Butene (highlighted red)

trans-2-Butene (highlighted blue)

5 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.

scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg

Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0

4 years ago

Part A

den301095 [7]

Answer:

$n_{Cl_2}=0.3molCl_2$

Explanation:

Hello there!

In this case, according to the given chemical reaction whereas the sodium chloride is in a 2:1 mole ratio with chlorine, the required moles of the later are computed as shown below:

$n_{Cl_2}=0.6molNaCl*\frac{1molCl_2}{2molNaCl}$

So we cancel out the moles of NaCl to obtain:

$n_{Cl_2}=0.3molCl_2$

Best regards!

3 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$