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DIA [1.3K]
3 years ago
11

The diagram above shows the results of Rutherford’s experiment in which he used a radioactive source to “shoot” alpha particles

at a thin sheet of gold foil. Based on these results, what were Rutherford’s conclusions?
A. Atoms are solid matter with positive and negative charges scattered throughout.
B. Atoms are mostly empty space with small, dense, negatively charged centers.
C. Atoms are solid, positively charged matter with negatively charged electrons scattered throughout.
D. Atoms are mostly empty space with small, dense, positively charged centers.
Chemistry
1 answer:
satela [25.4K]3 years ago
3 0

D is the answer its Atoms are mostly empty space with small, dense, positively charged centers.

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When nitrogen dioxide gas dissolves in water, an aqueous solution containing dissolved nitric acid and nitrogen monoxide forms.
xeze [42]

Answer:

1. Based on the given question, 138.03 grams of NO₂ is reacting completely with 18.02 grams of H2O. However, in case when 359 grams of NO₂ is used then the grams of water consumed in the reaction will be,  

= 359 × 18.02 / 138.03 = 46.87 grams of water.  

2. As mentioned in the given case 138.03 grams of NO₂ generates 126.04 grams of HNO₃. Therefore, 359 grams of NO₂ will produce,  

= 359 × 126.04 / 138.03  

= 327.81 grams of HNO₃.  

3. Based on the given question, 138.04 grams of NO₂ is generating 30.01 grams of NO. Therefore, 359 grams of NO₂ will generate,  

= 359 × 30.01 / 138.04

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5 0
3 years ago
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4 0
2 years ago
The Nernst equation at 20oC is:
saw5 [17]

Answer:

a. -58 millivolts

Explanation:

The given Nernst equation is:

E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}

The equilibrium potential given by the Nernst equation can be determined by using the formula:

E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}

where:

gas constant(R) = 8.314 J/K/mol

Temperature (T) = (20+273)K

= 298K

Faraday constant F = 96485 C/mol

Number of electron on Cl = -1

E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}

E_{Cl^-} = - 0.05814  \ volts

\mathsf{E_{Cl^-} = - 0.05814  \times 1000 \  milli volts}

\mathsf{E_{Cl^-} \simeq - 58\   milli volts}

5 0
3 years ago
Sn + 2Ha -> SnCl2 + H2
drek231 [11]

Answer:

your question is totally wrong. there is no cl2 in reactants.

4 0
3 years ago
How many moles are in 5.32 x 1020 atoms<br> of copper?<br> What is the answer but in numbers ?
Wewaii [24]

Answer is 1.41 x 10 with the exponent of 24, atoms

Explanation : Look at the picture i attached.

5 0
3 years ago
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