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cestrela7 [59]
3 years ago
12

From this list, choose all of the ionic compounds. Check all that apply. View Available Hint(s) Check all that apply. POCl3 KOCH

2CH3 CH3OH SOCl2 CH3CH2CO2Na NaOH CH2
Chemistry
1 answer:
Alik [6]3 years ago
4 0

Answer: KOCH_2CH_3, CH_3CH_2COONa, NaOH are all ionic compounds.

Explanation:

It is known that ionic compounds are the compounds in which one atom transfer its valence electrons to another atom. Hence, during this transfer partial opposite charges tend to develop on the combing atoms due to which strong force of attraction exists between the atoms.

An ionic bond is always formed between a metal and a non-metal.

For example, KOCH_2CH_3, CH_3CH_2COONa, NaOH are all ionic compounds.

On the other hand, a compound formed due to sharing of electrons between the combining atoms is known as a covalent compound. Generally, a non-metal with same or different non-metal tends to form a covalent bond.

For example, POCl_{3}, SOCl_{2} etc are all covalent compounds.

Thus, we can conclude that KOCH_2CH_3, [tex]CH_3CH_2COONa, NaOH are all ionic compounds.

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What is the mass percent of oxygen in sodium bicarbonate (NaHCO3)?
Alenkasestr [34]
The answer is 57.14%.

First we need to calculate molar mass of <span>NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.

Relative atomic mass of Na is 22.99 g
</span><span>Relative atomic mass of H is 1 g
</span><span>Relative atomic mass of C is 12.01 g
</span><span>Relative atomic mass of O is 16 g.
</span>
Molar mass of <span>NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 </span>· <span>16 g = 84 g

Now, mass of oxygen in </span><span>NaHCO3 is:
3 </span>· 16 g = 48 g

mass percent of oxygen in <span>NaHCO3:
48 g </span>÷ 84 g · 100% = 57.14%

Therefore, <span>the mass percent of oxygen in sodium bicarbonate is 57.14%.</span>
8 0
3 years ago
5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H
cricket20 [7]
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572.  That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

7 0
3 years ago
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I believe it is pentane..Try that!


Hope it helps, sorry if it doesn't but that's the answer I just choose for my unit test on edgnty.

5 0
3 years ago
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