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3 years ago

titration 25.0 ml of 3.4 x 10^ -3 M of Ba(OH) Neutralize 16.6 mL of HCL solution. What is the molarity of HCL solution?

1 answer:

8 0

What we are give: Concentration of base (CB) = 3.4 × $10^{-3}$
Then convert all volume in ml to L.
Volume of base (VB) 25.0ml = 0.025L
Volume of acid (VA) 16.6ml = 0.0166L
Now that we have everything we use the formula CAVA=CBVB.
Make 'CA' the subject then solve.
CA= $\frac{CBVB}{VA}$

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Find δs∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia:12n2(g) 32h2(g)→nh3(g)

Dima020 [189]

Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).

What is standard entropy?

The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."

Calculation:

Balancing the given reaction following-

1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)

ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]

Here S° = standard entropy of the system

Insert into the aforementioned equation all the typical entropy values found in the literature:

ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]

⇒ΔS° = - 99.4 J/K

Therefore, the standard entropy, ΔS° is -99.4 J/K.

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5 0

1 year ago

PLEASE HELP! Count the total number of atoms in H 2 O: 2 3 4 5 6

Darya [45]

3
two hydrogen and 1 oxygen

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3 years ago

Read 2 more answers

Calculate the value deltaG°

atroni [7]

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

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3 years ago

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

Fiesta28 [93]

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

$N= N_0(\frac{1}{2})^n$

where

$n = \frac{t}{t_{\frac{1}{2}}}$

here

t = 458,000 years

$t_{\frac{1}{2}}$ = 229,000

$\frac{t}{t_{\frac{1}{2}}}$ = \ $\frac{ 458,000}{229,000}$

n = $\frac{t}{t_{\frac{1}{2}}}$ = 2.000

Now substituting the values

$1.675 = N_0(\frac{1}{2})^{2.000}}$

$1.675 = N_0\times (0.2500)$

$N_0= \frac{1.675}{0.2500}$

$N_0=6.70$

3 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

3 years ago