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2 years ago

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:

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)

Explanation

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.

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3 years ago

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Question 3 of 8

vitfil [10]

Answer:

Explanation:

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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 $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating 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 $H_2So_4$ produced:

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

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:

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

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3 years ago