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Mashutka [201]
3 years ago
13

How are zinc and sodium alike?

Chemistry
1 answer:
mamaluj [8]3 years ago
5 0

Answer: both are solid at room temperture

Explanation:

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* Who can help with this lol *
Viktor [21]

Answer: subduction and sea floor spreading

Explanation:

He knew that the continents today were once joined together by fossil records of plants and animals that were found to be on continents far removed from each other. He knew this also by corresponding land forms that matches up as well. What he couldn’t prove is how the land masses would have moved so far away from each other. Subduction and sea floor spreading move the tectonic plates that the continents sit on. That’s what he was missing.

5 0
3 years ago
Simple Chem question - help!
Lunna [17]

d, one atom of oxygen and two atoms of hydrogen

6 0
3 years ago
i'm reasking this, i need help with this and it's due today. the answer isn't 30,240 minutes, it's the blanks i need filled in :
adelina 88 [10]

Answer:

Explanation:

24hours ×21 days=

you will get the answeer

3 0
3 years ago
G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu
Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄Ag^{+} + Br^{-1}.

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains \frac{0.6485}{45}×1000 = 14.411 gm-mole. Thus the solubility product K_{sp}of AgBr = [Ag^{+}]×Br^{-}.

Or, 5.0×10^{-13} = S^{2} (the given value of solubility product of AgBr is 5.0×10^{-13} and the charge of the both ions are same).

Thus S = (5.00×10^{-13})^{1/2} = 7.071×10^{-7} g/mL.

Thus the concentration of Br^{-1} or HBr is 7.071×10^{-7} g/mL.

4 0
3 years ago
The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer
sergey [27]

Answer:

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Explanation:

The pH of a buffer solution is calculated using following relation

pH=pKa+log(\frac{salt}{acid} )

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.

pKa = -log [Ka]

For HC₃H₅O₃

pKa = 3.1

For CH₃NH₃⁺

pKa = 10.64

pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

8 0
3 years ago
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