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2 years ago

Help me for a hefty reward

Chemistry

1 answer:

Mekhanik [1.2K]2 years ago

7 0

Group B is incorrect,They most likely added too much acid .

Let's check the reaction

$\\ \tt\hookrightarrow HCl+NaOH=NaCl+H_2O$

HCl is a strong acid
NaOH is a strong base
So it will lead to neutralization reaction.

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Which statement describes a chemical reaction?

ELEN [110]

The statement that describes a chemical reaction is the statement c. reactants form products when atoms are rearranged. The atoms of the reactants are arranged is a specific position and number forming a determined compound. In a chemical reaction, the products are different compounds than the reactants, so the atoms have to rearrange to form these new compounds (the products).

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3 years ago

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PLEASE HELP FAST!! How much energy (in calories) is need to melt 3.0 moles of ice?

Viktor [21]

Answer:

Energy

Explanation:

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Q= mL (Q= Heat required, m= mass, L= Latent Heat of Fusion/Evaporation/Anything else).

So, according to your values:

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3 years ago

Which of the following occupies a volume? solids liquids gases all of the above

nataly862011 [7]

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3 years ago

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What is the difference between a hydrogen ion and a hydroxide ion?

Sophie [7]

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there is some difference

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hydrogen ion --

symbol-H

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symbol- OH

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they are bonded.

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5 0

3 years ago

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:

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

5 0

3 years ago