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Natasha_Volkova [10]
3 years ago
10

What does the first word in the name of a binary ionic compound represent?

Chemistry
2 answers:
Sauron [17]3 years ago
6 0

Answer : The correct option is, The positive ion

Explanation :

The rules for naming the ionic compounds is given as :

  • The positive ion (cation) is written first.
  • The negative ion (anion) is written next.
  • The suffix is added at the end of the negative ion (anion). The suffix used is '-ide'.

As per question, the first word in the name of a binary ionic compound represent the positive ion and then the negative ion.

For example : NaCl is the binary ionic compound that composed of two different elements which are Na and Cl. The name of this compound is, Sodium chloride.

Hence, the correct option is, The positive ion

Otrada [13]3 years ago
4 0

Answer:

The positive ion

Explanation:

Took the test and got it correct.

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of
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Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

mol=0.282*0.01

mol=0.00282

We can do the same for the KOH values (50mL=0.05L and 0.141M).

0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

0.00705-0.00282=0.00423mol~of~KOH

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Total~volume=50mL+10mL=60mL

60mL=0.06L

Now we can calculate the concentration:

M=\frac{0.00423mol}{0.06L}

M=0.0705

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

pOH=-Log(0.0705)

pOH=1.15

Now we can <u>calculate the pH value</u>:

14=~pH~+~pOH

pH=14-1.15=12.84

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D. 20/10 g/ml

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