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babunello [35]
3 years ago
14

The element thallium is 70% thallium-205 and 30% thallium-203. Calculate its relative atomic mass to 1 decimal place.

Chemistry
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

answer-

The relative atomic mass = 204.4

explanation:

Thallium -203 = 30%

Thallium -205 = 70%

Therefore ,

relative mass of thallium = (30×203 + 70×205)/100

relative mass of thallium = (20440)/100

relative mass of thallium = 204.40 amu

Thus,

relative atomic mass of thalium =204.4 ( to 1 decimal place)

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A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>

1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

3 0
3 years ago
The mass percent of solute in a solution containing 3.73 g KBr dissolved in 131 g of H2O is:
Elina [12.6K]

\huge \underbrace \mathfrak \red{Answer}

28%

Explanation:

mass of solute(KBr) = 3.73g

mass of solvent(H2O) = 131g

mass of solution = mass of solute + mass of solvent

= 3.73 + 131

= 134.73g

\sf \large {mass \: percentage =  \frac{mass \: of \: solute}{mass \: of \: solvent}  \times 100} \\  \\  \sf  mass \: percentage =  \frac{3.73}{134.73}  \times 100 \\  \\  \sf mass \: percentage =  0.028 \times 100 \\  \\  \sf mass \: percentage = 28\%

7 0
2 years ago
Explain and describe how the photoelectric effect occurs on an atomic level in terms of protons, neutrons, and electrons
grin007 [14]

Answer:

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons.

Based on the wave model of light, physicists predicted that increasing light amplitude would increase the kinetic energy of emitted photoelectrons, while increasing the frequency would increase measured current.

3 0
3 years ago
Salts are which part of an aqueous solution?
masha68 [24]

Explanation:

Salts are the solutes in an aqueous solution. An aqueous solution is solution whose solvent water.

  • To form a solution,a solute must be dissolved in a solvent.
  • For aqueous solutions, the solvent which is the dissolving medium is made up of water.
  • The solute is the substance that is dissolved in it.
  • Salts for example can be a solute in an aqueous solution.
  • A salt is generally an ionic compound consisting of positive ions such as metallic, ammonium ans complex ions and negative ions such as acid radicals and complex ions.

Learn more:

Aqueous solution brainly.com/question/8426727

#learnwithBrainly

3 0
3 years ago
When 150. g zinc sulfide are burned in excess oxygen, 68.5 g of zinc oxide are actually produced, along with sulfur dioxide. Det
Bumek [7]

%yield = 54.6%

<h3>Further explanation</h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

(theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

<h3 />

Reaction

2ZnS+3O₂ ⇒ 2ZnO+2SO₂

MW ZnS = 97.474 g/mol

  • mol ZnS

\tt \dfrac{150}{97.474}=1.54

MW ZnO = 81.38 g/mol

  • mol ZnO (from mol ZnS as limiting reactant, O₂ excess)

\tt \dfrac{2}{2}\times 1.54=1.54

  • Actual ZnO produced

\tt 1.54\times 81.38=125.33~g

Theoretical production = 125.388

  • %yield

\tt \dfrac{68.5}{125.33}\times 100\%=\boxed{\bold{54.6\%}}

4 0
3 years ago
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