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Aloiza [94]
2 years ago
15

2 multiple choice questions, Please help

Chemistry
1 answer:
gizmo_the_mogwai [7]2 years ago
6 0

1) b. Radioactive

2. a) tevatron

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Knowledge of atomic structure has evolved over time. Thomson discovered that atoms contain electrons, Rutherford discovered that
sergejj [24]
I believe the correct response would be A. The cumulative nature of science. How experimental evidence and ideas made by many scientists have accumulated and or built up to have a more understandable and or clear view of a particular scientific topic, and or principle.
4 0
3 years ago
Read 2 more answers
A box contains identical balls of which 12 are red, 18 white and 8 blue. Three balls are drawn from the box one after the other
svet-max [94.6K]

Answer:

a) 12/323

b) 8/233

Explanation:

a) The probability of a red ball being drawn is 12/38, or in a simplified fraction, 6/19. To find the probability that 3 are red you would multiply the probability of the fraction for each, except subtracting one from the total each time as the drawn is done without replacement. This is done as follows: 6/19 × 6/18 × 6/17= 12/323

b) The probability of drawing a blue ball is 8/38, or 4/19. To find that the first one is blue and the rest are red, the equation is done as follows: 4/19 × 6/18 × 6/17 = 8/233

(hopefully I did this right)

8 0
3 years ago
A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
Paul [167]

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

5 0
3 years ago
Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.
o-na [289]

Answer:

vHe / vNe = 2.24

Explanation:

To obtain the velocity of an ideal gas you must use the formula:

v = √3RT / √M

Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio is:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

<em>vHe / vNe = 2.24</em>

<em />

I hope it helps!

8 0
3 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
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