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zhuklara [117]
3 years ago
9

Match the characteristics and elements to their group.

Chemistry
1 answer:
Luba_88 [7]3 years ago
8 0

Answer:

Kr is a Noble Gas. Na is an alkali metal. F is halogen.

Group 17 is halogens. Inert is Noble Gases. Odourless and colourless is Noble Gases. Alkali metals do not occur freely in nature. Alkali metals are malleable

Explanation:

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On the first day of your new job as a chemist, you are given a bottle of magnesium sulfate and asked to make 30 mL of 0.3 M MgSO
Lady bird [3.3K]

Answer:

The answer to your question is 15500 mg of MgSO₄ 7H₂O

Explanation:

Data

Volume = 30 ml

Concentration = 0.3 M

Formula MgSO4∗7H2O

Process

1.- Calculate the number of moles needed

 Molarity = moles / volume

Solve for moles

 moles = Molarity x volume

Substitution

 moles = 0.3 x 0.03 l

 moles = 0.009 moles of MgSO₄

2.- Calculate the molecular mass of MgSO4∗7H2O

molecular mass = 24 + 32 + 64 + 14 + 112

                           = 246 g

3.- Get the proportion MgSO4 :7H2O, this proportion is 1 : 7

4.- Calculate the amount of MgSO4∗7H2O

    0.009 mol of MgSO4 \frac{7 MgSO4 7H2O}{1 mol MgSO4} \frac{246 g MgSO4 7H2O}{1 mol MgSO4 7H2O}

Simplification

   15.5 g = 15500 mg

5 0
3 years ago
Write the IUPAC name OF THE FOLLOWING COMPOUNDS
goldenfox [79]

Answer:

Explanation:

a) 2 chloro butane

b) 2-3 dimethyl butane

c) 2 bromo 3 nitro pentane

d) 2-3 trimethyl pentane

e) 2-bromo,3-methyl,4-nitro hexane

f) 2-methyl cyclo butane

7 0
3 years ago
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
What did J.J. Thomson discover about the atom that changed the atomic model previously used?
galina1969 [7]
J.J. Thomson hypothesized and discovered that the atom was not the smallest unit of matter but that instead there were much smaller units. He discovered "sub-atomic particles" which make up atoms. The sub-atomic particle that Thomson discovered was the electron. He discovered this through a process of experiments testing cathode rays.
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3 years ago
Could a mixture be made up of only elements and no compounds. explain
lbvjy [14]
Yes, because a mixture is 2 or more substances that are mixed together (not chemically). A mixture could be two different elements physically combined in a set ratio.
8 0
3 years ago
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