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1 year ago

based on distance to the nearest water well, which drilling site would pose the least risk of methane contamination?

1 answer:

7 0

Based on distance to the nearest water well, the drilling site that would pose the least risk of methane contamination would be the one farthest away from the water well.

Methane is a gas that is known to escape from the ground and rise up into the atmosphere. If the drilling site is too close to the water well, there is a greater risk that the methane could contaminate the water.

The atmosphere is the air that surrounds us. It is a mixture of gases, including nitrogen, oxygen, and carbon dioxide. The atmosphere protects us from the Sun's ultraviolet radiation and from the cold of space. It also contains the Earth's weather, which is caused by the Sun's energy.

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Which represents a balanced nuclear equation?

shtirl [24]

Answer: second answer

Explanation:

The thing is that every new nuclear cycle a new element forms and reduces an electron. I may be wrong but this is the most logical and scientifically correct answer

3 0

3 years ago

If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

<em> </em>inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$

Therefore, the new pH is 7.69.

Learn more about pH here:

brainly.com/question/24595796

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8 0

2 years ago

Atoms are bonded together in a different order in

Savatey [412]

Answer:

answer.A

Explanation:

that is my answer

4 0

2 years ago

Read 2 more answers

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.