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Ray Of Light [21]
3 years ago
12

Which of the following statements about functional groups is false? Group of answer choices The only functional group possible i

n a hydrocarbon is the double bond. An alkane does not have a functional group. The functional group of an alcohol is the –OH group. The functional group represents the most common site for reactivity. The double bond is the functional group of an alkene.
Chemistry
1 answer:
Alborosie3 years ago
4 0

Answer: the false statement is:

The only functional group possible in a hydrocarbon is the double bond

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List the number of each type of atom on the right side of the equation hbr(aq)+2naoh(aq)→2nabr(s)+h2o(l)
ser-zykov [4K]

On the basis of the given unbalanced equation, that is:

HBr (aq) + 2NaOH (aq) → 2NaBr (s) + H2O (l)

On the right side of the equation, there are 2 atoms of sodium (Na), 2 atoms of bromine (Br), 2 atoms of hydrogen (H), and 1 atom of oxygen (O₂).

After balancing the equation correctly we get:

HBr (aq) + NaOH (aq) → NaBr (s) + H2O (l)

On the right side, one atom of Na, 1 atom of Br, 1 atom of H and one atom of O₂.

5 0
3 years ago
Read 2 more answers
How many mL of 0.506 M HClare needed to dissolve 9.85 g of BaCO3?
Blababa [14]

Answer:

197mL of 0,506M HCl

Explanation:

The reaction of HCl + BaCO₃ is:

BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.

The moles of BaCO₃ in 9,85 g are:

9,85 g of BaCO₃ × \frac{1mol}{197,34 g} = <em>0,0499 moles of BaCO₃</em>

As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:

0,0499 moles of BaCO₃ × \frac{2molHCl}{1molBaCO_{3}} =<em> 0,0998 moles of HCl</em>

If you have a 0,506M HCl, you need to add:

0,0998 moles of HCl× \frac{1L}{0,506moles} = 0,197 L ≡ 197mL

I hope it helps!

8 0
3 years ago
Atoms of the element beryllium would most likely behave similar to the way _______ behaves.
lubasha [3.4K]
It's b. calcium 
because the outer electron structure in all of that element is similar, they have the same chemical and physical properties.<span />
3 0
3 years ago
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At what point is base no longer added during a titration
maria [59]
When you doing a titration, you need to use an indicator to confirm whether the reaction is completed. When the indicator has the color change and will not change back in one minute, the reaction is finished and you don't need to add more.
8 0
3 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0
3 years ago
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