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3 years ago

How much faster does nitrogen monoxide (NO) effuse in comparison to dinitrogen tetraoxide (N2O4)?

Chemistry

1 answer:

juin [17]3 years ago

5 0

Answer:

N5O2

Explanation:

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Bromine has two isotopes whose masses are 78.9183 amu and 80.9163 amu. Their abundances are 50.69% and 49.31%, respectively. Cal

artcher [175]

Answer:

79.9035

Explanation:

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3 years ago

How much NaCl would you dissolve and dilute to the 100.00 mL mark in a volumetric flask with DI H2O to prepare a 0.825 M sodium

adoni [48]

Answer:

4.82 g

Explanation:

To solve this problem we'll use the <em>definition of molarity</em>:

Molarity = moles / liters

We are given the volume and concentration (keep in mind that 100mL=0.100L):

0.825 M = moles / 0.100 L

Now we <u>calculate the number of NaCl moles required</u>:

moles = 0.0825 mol

Then we convert 0.0825 NaCl moles into grams, using its molar mass:

0.0825 mol * 58.44 g/mol = 4.82 g

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3 years ago

What are the factors affecting evaporation?

vazorg [7]

Answer:

Temperature: The rate of evaporation increases with an increase in temperature. Surface area: The rate of evaporation increases with an increase in surface area. Humidity: The amount of water vapour present in the air is called humidity.

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10. A 20.00 mL sample of 0.150 mol/L ammonia (NH3(aq)) is titrated to the equivalence point by 20.0 mL of a solution of 0.150 mo

Natalija [7]

Answer:

$\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}$

Explanation:

a) Balanced equation

The balanced chemical equation for the titration is

$\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq)}$

b) pH at start

For simplicity, let's use B as the symbol for NH₃.

The equation for the equilibrium is

$\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}$

(i) Calculate [OH]⁻

We can use an ICE table to do the calculation.

B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹: 0.150 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.150 - x x x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150$

(ii) Solve for x

$\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}$

(iii) Calculate the pH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}$

(c) pH at equivalence point

(i) Calculate the moles of each species

$\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}$

B + HI ⇌ BH⁺ + I⁻

I/mol: 3.00 3.00 0

C/mol: -3.00 -3.00 +3.00

E/mol/: 0 0 3.00

(ii) Calculate the concentration of BH⁺

At the equivalence point we have a solution containing 3.00 mmol of NH₄I

Volume = 20.00 mL + 20.00 mL = 40.00 mL

$\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}$

(iii) Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

BH⁺+ H₂O ⇌ H₃O⁺ + B

I/mol·L⁻¹: 0.0750 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.0750 - x x x

$K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ $\ll$ 0.0750}$

$\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}$

(iv) Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}$

The titration curve below shows the pH at the beginning and at the equivalence point of the titration.

8 0

3 years ago