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3 years ago

When an unknown substance is dissolved in water, hydronium ions form. What can you conclude about the substance? @thomaster

1 answer:

5 0

HIdronium ions, H3O +, are formed as consequence of protonation of water.

The protons that cause this protonation come from acids.

Then, you can infere that the substance is an acid.

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What<br>was the initial volume of the hydrogen in cm3?

svetlana [45]

Answer:

255.51cm3

Explanation:

Data obtained from the question include:

V1 (initial volume) =?

T1 (initial temperature) = 50°C = 50 + 273 = 323K

T2 (final temperature) = - 5°C = - 5 + 237 = 268K

V2 (final volume) = 212cm3

Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:

V1/T1 = V2/T2

V1/323 = 212/268

Cross multiply to express in linear form

V1 x 268 = 323 x 212

Divide both side by 268

V1 = (323 x 212)/268

V1 = 255.51cm3

Therefore, the initial volume of the gas is 255.51cm3

5 0

3 years ago

Read 2 more answers

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Dmitrij [34]

Answer:

No hay resultados para ACFrOgDjZVP5d1mO2l6HAtPZQNnbFRq674g04E7uZadqJMPc4VbhdTIEDCWBeh3xfw9BrKkfHHEN4nxe9NVglsb9N8D49CjxvxHYw3L93m4wO6SY5SwKQYMk-2zzHtGzerun1Uh9k-mpxFw9D3I1.pdf

Explanation:

hablemos por g mail

5 0

3 years ago

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.29 mol of m

gregori [183]

Answer:

1.645 moles of excess reactant that is of magnesium metal are left over.

Explanation:

Moles of magnesium metal = 3.29 mol

Moles of HCl = 3.29 mol

$Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)$

According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :

$\frac{1}{2}\times 3.29 mol=1.645 mol$ of magnesium metal

Moles of HCl left = 3.29mol - 3.29 mol = 0

Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol

1.645 moles of excess reactant that is of magnesium metal are left over.

7 0

3 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

4 years ago

What two particles must a substance particles exhibit in order to conduct electricity

Tom [10]

Answer:

electron (-) and proton (+)

5 0

3 years ago