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Gre4nikov [31]
3 years ago
11

suppose 100ml of perchloric acid is neutralized by exacly 46.9 ml of 0.75m lithium hydroxide.what is the concentration of the pe

rchloric acid?​
Chemistry
1 answer:
allochka39001 [22]3 years ago
6 0

Answer:

0.352M

Explanation:

First, we need to write a balanced equation for the reaction between perchloric acid and lithium hydroxide. This is shown below:

HClO4 + LiOH —> LiClO4 + H2O

From the balanced equation above, we obtained:

nA (mole of the acid) = 1

nB (mole of the base) = 1

Data obtained from the question include:

Va (volume of the acid) = 100mL

Ca (concentration of the acid) =?

Vb (volume of the base) = 46.9 mL

Cb (concentration of the base) = 0.75M

Using CaVa/CbVb = nA/nB, the concentration of the acid can obtained as follow:

CaVa/CbVb = nA/nB

Ca x 100 / 0.75 x 46.9 = 1

Cross multiply to express in linear form

Ca x 100 = 0.75 x 46.9

Divide both side by 100

Ca = (0.75 x 46.9) /100

Ca = 0.352M

Therefore the concentration of the acid is 0.352M

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likoan [24]

Answer:

3 significant numbers

6 0
3 years ago
The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J
Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

<em>A = 1,13x10¹⁰</em>

<em></em>

I hope it helps!

6 0
2 years ago
A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man
max2010maxim [7]

Answer:

There will react 0.400 moles of oxygen.

Explanation:

<u>Step 1:</u> Data given

Volume of the closed flask = 200.00 mL = 0.2 L

Number of moles of CO = 2.000 mol

Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

<u>Step 2</u>: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

<u>Step 3</u>: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

<u>Step 4:</u> When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

<u>Step 5:</u> Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

⇒ V = the volume = 200.0 mL = 0.2 L

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

n = 3.6 moles = total number of moles

<u>Step 6:</u> Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

Moles O2 = (2-X) moles

Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

6 0
3 years ago
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is starte
Katarina [22]

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

3 0
3 years ago
How to balance this equation in chemistry
saveliy_v [14]

Answer:

4NH₃(g) +3O₂(g) ⇒2N₂(g) +6H₂O(g)

Explanation:

4 0
2 years ago
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