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MariettaO [177]
3 years ago
10

What type of bonds are these?

Chemistry
1 answer:
gulaghasi [49]3 years ago
3 0
Jeez this is middle school really. covalently bond
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375 grams

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Zelda noticed a puddle outside her front door. She saw that the puddle got smaller every day, mil the
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to set 1 container inside, without air movement. 1 outside in that location. compare the 2 containers to see which container has less or more fluid...

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Which best represents the reaction of calcium and zinc carbonate (ZnCO3) to form calcium carbonate (CaCO3) and zinc? Ca → ZnCO3
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When the pressure on 2.5 liters of oxygen is decreased from 2.0 atm to 1.0 atm, the volume of the gas
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Answer:

volume of the gas is 5.0L

Explanation:

Using Boyle's law that state the pressure of a gas is inversely proportional to volume of it occupies when temperature is constant, it is possible to write:

P₁V₁ = P₂V₂

<em>Where P is pressure, V is volume and 1 and 2 are initial and final states.</em>

<em />

If initial volume is 2.5L, initial pressure is 2.0atm and 1.0atm is final pressure, final volume is:

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5.0L = V₂

Thus, <em>volume of the gas is 5.0L</em>.

5 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

#SPJ1

 

8 0
1 year ago
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