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Inessa05 [86]
2 years ago
7

What mass of sucrose is needed to make 300.0 mL of a 0.5 M solution? (molar mass= 342.34 g/mol)

Chemistry
1 answer:
Anuta_ua [19.1K]2 years ago
7 0
I need points i’m sorry but yea lol good luck
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Ability to moderate temperatures is not a property of water

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2 years ago
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Help it’s about Binary compounds and transition metals
Musya8 [376]

Answer:

Cr 6+ & SO4 2-

Explanation:

Sulfate is a polyatomic ion that has a charge of 2-. There are three of them in the chemical formula so it equates to a 6- total charge. Thus, chromium must have a 6+ charge to give the compound a neutral charge.

8 0
2 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
This is science help me pls
jek_recluse [69]
Remember velocity = distance/time so that mean you would do 60/3 which equals 20 so the speed(velocity) equals 20
6 0
1 year ago
Which models of the atom in task 1 are not supported by the results of the hydrogen gas experiment? For each of these models, ex
kramer

Answer:

Thomson  placed two magnets on either side of the tube, and observed that this magnetic field also deflected the cathode ray. The results of these experiments helped Thomson determine the mass-to-charge ratio of the cathode ray particles, which led to a fascinating discovery, minus the mass of each particle was much, much smaller than that of any known atom. Thomson repeated his experiments using different metals as electrode materials, and found that the properties of the cathode ray remained constant no matter what cathode material they originated from. From this evidence, Thomson made the following conclusions:

The cathode ray is composed of negatively-charged particles.

The particles must exist as part of the atom, since the mass of each particle is only ~1/2000 the mass of a hydrogen atom.

These subatomic particles can be found within atoms of all elements.

While controversial at first, Thomson's discoveries were gradually accepted by scientists. Eventually, his cathode ray particles were given a more familiar name: electrons. The discovery of the electron disproved the part of Dalton's atomic theory that assumed atoms were indivisible. In order to account for the existence of the electrons, an entirely new atomic model was needed.

Explanation:

8 0
2 years ago
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