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Darina [25.2K]
3 years ago
12

Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same

voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 1.67×10^{-27} kg.
Chemistry
1 answer:
Tju [1.3M]3 years ago
5 0

Answer:

42.8

Explanation:

A particle travelling through potential difference V has energy E =qV = 1/2mv²

v = √(2qV / m)

ve = √(2qV / me)

vh = √(2qV / mh)

ve / vh = √( (2qV / me) ÷(2qV / mh) = √ ( mh / me ) where mh = mass of hydrogen ion and me is mass of electron, V voltage is the same, ve is speed of electron and vh is the speed of hydrogen ion

ve / vh  = √ ( mh / me )  = √ ( 1.67 × 10⁻²⁷ kg) / ( 9.11 × 10 ⁻³¹ kg) = 42.8

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Please help asap and thank you
natulia [17]

Answer:

Metamorphic rock

Explanation:

metamorphic rocks do not get hot enough to melt they just get denser if they did they would become igneous rocks.

7 0
3 years ago
A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif
noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = [A_o]

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = [A]=0.06[A_o]

Half life of the carbon-14 = t_{1/2}=5730 years

k=\frac{0.693}{t_{1/2}}

[A]=[A_o]\times e^{-kt}

[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}

0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0
3 years ago
Comment Both propane and benzene are hydrocarbons. As a rule,
kozerog [31]

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

5 0
3 years ago
A helium balloon with an internal pressure of 1atm and a volume of 4.20 L at 18.0°C is released. What volume will the balloon oc
kari74 [83]

Answer:

7.59 L

Explanation:

- Use combined gas law formula and rearrange.

- Change C to K

- Hope that helped! Please let me know if you need further explanation.

3 0
3 years ago
G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu
Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄Ag^{+} + Br^{-1}.

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains \frac{0.6485}{45}×1000 = 14.411 gm-mole. Thus the solubility product K_{sp}of AgBr = [Ag^{+}]×Br^{-}.

Or, 5.0×10^{-13} = S^{2} (the given value of solubility product of AgBr is 5.0×10^{-13} and the charge of the both ions are same).

Thus S = (5.00×10^{-13})^{1/2} = 7.071×10^{-7} g/mL.

Thus the concentration of Br^{-1} or HBr is 7.071×10^{-7} g/mL.

4 0
3 years ago
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