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3 years ago

Where do the OH- ions go as additional acid is added?

2 answers:

8 0

The result when the OH- ions go as additional acid is added is that they remain in solution but change concentration. This is because when you add more acid to hydroxide, it will change its concentration due to the acidity added.

stiks02 [169]3 years ago

8 0

Answer:

A: They evaporate

Explanation:

Where do the OH- ions go as additional acid is added? They evaporate. They combine with excess H+ ions to form more water. They remain in the solution at the same concentration.

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When converting an acid salt dissolved in water to its acid form, you are instructed to adjust the ph well into the acidic range

777dan777 [17]

First of all, the problem says that you have to convert the acid salt to its acidic form. If you take it to the neutral pH, that won't be acidic at all. As simple as that, you don't take it to neutral pH because it would lose its definition of being acidic afterall.

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3 years ago

What is the mass of 14.4 moles Fluorine? 0.379g 274.8 1.32 g 2.64g 547 g

liberstina [14]

Answer:

Explanation: d

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3 years ago

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Dd is recessive or dominant

slava [35]

Answer:This would be heterozygous, so both dominant and recessive alleles are written.

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3 years ago

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Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

$V_s=V_o-V_i=V_i\\V_o=2*V_i\\$

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

$r_o=r_i+e$

The first equation becomes

$\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0$

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

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2 years ago

Help me please ( I'll give brainlist to whoever answers right)

devlian [24]

A goes with u, C with G, and T with A

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2 years ago