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3 years ago

If .758 moles of gas occupy a volume of 80.6L, how many moles will occupy a volume of 270.9L?

Chemistry

1 answer:

egoroff_w [7]3 years ago

4 0

Answer:

n₂ = 2.55 mol

Explanation:

Given data:

Initial number of moles = 0.758 mol

Initial volume = 80.6 L

Final volume = 270.9 L

Final number of moles = ?

Solution:

Formula:

V₁/n₁ = V₂/n₂

V₁ = Initial volume

n₁ = initial number of moles

V₂ = Final volume

n₂ = Final number of moles

now we will put the values in formula.

V₁/n₁ = V₂/n₂

80.6 L / 0.758 mol = 270.9 L/ n₂

n₂ = 270.9 L× 0.758 mol / 80.6 L

n₂ = 205.34 L.mol /80.6 L

n₂ = 2.55 mol

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Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.

dexar [7]

8.8 × 10-5 M is the [H3O+] concentration in 0.265 M HClO solution.

Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

HClO +H2O ⇒ ClO- + H3O+

I 0.265 0 0

C -x +x +x

E 0.265-x +x +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

Ka = $\frac{[ClO-][H3O+]}{[HClO]}$

2.9 × 10^-8 = $\frac{[x] [x]}{[0.265-x]}$

$x^{2}$ = 7.698 x $10^{-9}$

x = 8.8 × 10-5 M

The hydronium ion concentration is 8.8 × 10-5 M in 0.265 M solution of HClO.

8 0

3 years ago

Perform the following

vivado [14]

The correct number of significant figures and digits is 3

5 0

3 years ago

Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider a mixture of six nitrogen molecules and six hydrogen molec

Nezavi [6.7K]

Answer:

a) No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

Explanation:

The balanced reaction between nitrogen and hydrogen molecules to give ammonia molecules is:

$N_{2}(g)+3H_{2}(g) -->2NH_{3}$

Thus one molecule of nitrogen will react with three molecules of hydrogen to give two molecules of ammonia.

We have six molecules of each nitrogen and hydrogen in the closed container and they undergo complete reaction it means the limiting reagent is hydrogen. For six molecules of nitrogen, eighteen molecules of hydrogen will be required.

So six molecules of hydrogen will react with two molecules of nitrogen to give four molecules of ammonia.

The product mixture will have

a) No molecules of hydrogen

b) four molecules of ammonia

c) four left molecules of nitrogen.

7 0

3 years ago

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

3 years ago

g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have

V125BC [204]

Answer:

pH = 2.66

Explanation:

Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid

1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

[H⁺] = $\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}$

[H⁺] = 0.0028

Finally we calculate the pH:

pH = -log[H⁺]

pH = 2.66

8 0

3 years ago