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leonid [27]
2 years ago
15

How many atoms of oxygen are in a molecule of glucose (C6H12O6)

Chemistry
2 answers:
valentinak56 [21]2 years ago
4 0

Answer:

It is 6

Explanation:

6 oxygen atoms

AveGali [126]2 years ago
4 0

Answer:

6 oxygen atoms

Explanation:

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Ierofanga [76]


Give any experiment you've learned so that I can explain thx.
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3 years ago
A white salt containing an unknown metal has the formula MCl and gives a lilac flame during a flame test. the salt could be
Tatiana [17]

Answer:

C

Explanation:

KCl - the flame test for Potassium produces a lilac flame

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2 years ago
Which of the following happens when chlorine forms an ion?
Jlenok [28]

What  happens  when  chlorine  form  an ion  is that  it gains an electron and  has  an octet  in its  outer shell  ( answer  A)


<u><em> Explanation</em></u>

<u><em> </em></u>Chlorine is  is in atomic  number  17  in periodic table.

The electron configuration  of chlorine  is    1S2 2S2 2P6 3S2 3P5   or[Ne]3S2 3p5  or  2.8.7.

chlorine therefore  has 7 valence electron therefore it  gain  1 electron  to form Cl- ( ion)

Cl- has  8  electron in its outer  shell (  it  obeys  octet  rule  of eight valence in outer shell.

7 0
3 years ago
Read 2 more answers
Question 3 of 8
vitfil [10]

Answer:

B

Explanation:

7 0
2 years ago
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The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
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