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lana66690 [7]
3 years ago
7

- The subatomic particle possessing the largest

Chemistry
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

B. electrons possess the largest charge-to-mass ratio among the subatomic particles listed in the four choices.

Explanation:

Consider the mass of each particle. Express the masses in atomic mass units:

  • Protons: approximately 1.007 amu each;
  • Neutrons: approximately 1.009 amu each;
  • Electrons: approximately 0.0005 amu each.

Similarly, consider the charge on each particle. Express the charges in multiples of the fundamental charge:

  • Protons: +1 e;
  • Neutrons: 0;
  • Electrons: -1 e.

Calculate the charge-to-mass ratio for the three species:

  • Protons: approximately \rm 0.99\; e\cdot amu^{-1};
  • Neutrons: 0;
  • Electrons: approximately \rm 1,800\;e \cdot amu^{-1}.

Almost all nuclei contain protons and neutrons. The only exception is the hydrogen-1 nucleus, which contains only one proton and no neutron. The mass of the nucleus is approximately the same as the sum of its components' masses. The extra neutron will only add to the mass of the nucleus (the denominator) without contributing to the charge (the numerator.) As a result, the charge-to-mass ratio of nuclei will be positive but no greater than the charge-to-mass ratio of protons.

Among the particles in the four choices, the charge-to-mass ratio is the greatest for electrons.

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Four cars travel at the same speed down a road. Their mass and kinetic energy are shown in the graph,
Alex777 [14]

Answer:B

Explanation:

3 0
2 years ago
in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
Organic Chem Rxn Question
NemiM [27]

Answer:

a, g, c

Explanation:

The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.

The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.

8 0
3 years ago
On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i
ddd [48]

Answer:

1.6 L

Explanation:

Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (32 + 273.15) K = 305.15 K  

Using above equation as:

\frac{1.5}{285.15}=\frac{V_2}{305.15}

V_2=\frac{1.5\cdot \:305.15}{285.15}

New volume = 1.6 L

8 0
3 years ago
In a particular redox reaction, no2– is oxidized to no3– and cu2 is reduced to cu . Complete and balance the equation for this r
lianna [129]

Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).


Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.  

Balanced chemical reaction:

Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).

Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).


8 0
3 years ago
Read 2 more answers
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