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almond37 [142]
3 years ago
7

How many milliliters of a stock solution of 6.20 M HNO3 would you have to use to prepare 0.130 L of 0.470 M HNO3?

Chemistry
1 answer:
Mandarinka [93]3 years ago
6 0

Answer:

9.85

Explanation:

M1V1 =M2V2

6.20×v1= 0.470×0.130

v1 = ( 0.470 × 0.130 ) ÷ 6.20

v1 = 0.0098 L × 1000

V1 = 9.8 ml

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Klio2033 [76]

Answer:

Atoms are the building blocks of matter. They are the smallest particles of an element that still have the element's properties. All atoms are very small, but atoms of different elements vary in size. Three main types of particles that make up all atoms are protons, neutrons, and electrons.

Explanation:

7 0
2 years ago
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Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu
borishaifa [10]

Answer:

\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

                    HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹:       2.7                   0       0

C/mol·L⁻¹:      -x                   +x      +x

E/mol·L⁻¹:   2.7 - x                 x        x

K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}

2. Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

     2.7                         x        0.0441

K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}

6 0
3 years ago
Please help me with this question!! I’m not sure if they’re the insoluble ions that precipitate with chromate or the soluble one
Vika [28.1K]

Answer:

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Explanation:

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4 0
2 years ago
How many moles are in 4177g of aluminum chloride
Luden [163]
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moles = 4177 g / </span><span>133.34 g/mol
moles = 31.33 moles</span>
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shtirl [24]

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Explanation:

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