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Svet_ta [14]
3 years ago
12

Solutions with low pH values have which of these properties?

Chemistry
2 answers:
telo118 [61]3 years ago
8 0

Answer:

The correct option is C.

Solutions with low pH values have sour taste.

Explanation:

Low pH value show acidic property which can be define as , -log of conc. of hydrogen ion.

Acid has following properties

  • Sour taste
  • Turn blue litmus into red
  • corrosive

-OH group: Base have OH groups. So it is not correct option so it not correct for low pH i.e. acid

turns litmus paper blue: Base turn red litmus into blue, so it not correct for low pH i.e. acid

feels slippery: Bases are mostly slippery like soap

tastes sour: Sour taste is for acid and bitter taste is for base. So it is correct option for acid

Sladkaya [172]3 years ago
7 0

Tastes sour

I did this test a while ago sorry if I'm wrong

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The question is incomplete, the complete question is shown in the image attached

Answer:

A and B

Explanation:

The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.

The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.

Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.

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How many moles (of molecules or formula units) are in each sample? 79.34 g cf2cl2?
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From the periodic table:
molecular mass of carbon = 12 grams
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Therefore, we can use cross multiplication to find the number of moles in 79.34 grams as follows:
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How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

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