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Kipish [7]
4 years ago
11

What experimental evidence led to the development of this atomic model

Chemistry
1 answer:
Yanka [14]4 years ago
8 0
Rutherford’s experimental evidence led to the development of this atomic model from the one before it . According to him , a few of the positive particles pointed at a gold foil appeared to ricochet back off of the light metallic foil. This experimental evidence gives the root the development of the atomic model .
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What is the oxidation number for each atom in NH4CI?
Alika [10]
<span>N = +3, H = +1 ,Cl = -1
</span><span>

</span>
5 0
4 years ago
For the reaction below, initially the partial pressure of all 3 gases is 1.0atm. . 2NH3(g)--&gt; N2(g) + 3H2(g) K, 0.83 1. When
erma4kov [3.2K]

Answer:

The reaction would shift toward the reactants

When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm

Explanation:

For the reaction:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Where K is defined as:

K = \frac{P_{N_{2}}*P_{H_2}^3}{P_{NH_3}^2} = 0.83

As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:

Q = \frac{1atm*{1atm}^3}{1atm^2} = 1

As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>

Thus, there are true:

<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>

<em />

4 0
3 years ago
Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units.
Thepotemich [5.8K]

Answer:

m = 2.955x10⁻² mol/kg

X = 5.323x10⁻⁴ mol NaCl/Total moles

(w/w)% = 0.1726%

ppm = 1726 mg/kg

Explanation:

Molality is the ratio between moles of solute per kg of solution.

As the solution is 2.950×10⁻² mol/L, mililters are 999.2mL and density is 0.9982 g/mL, molality is:

m = 2.950×10⁻² mol/L×(1L/0.9982kg) = <em>2.955x10⁻² mol/kg</em>

Mole fraction is moles of NaCl/total moles.

Moles of H₂O are:

999.2mL×(0.9982g/mL)×(1mol/18,02g) = 55,35 moles of H₂O

Moles of NaCl are:

2.950×10⁻² mol/L×(0.9992L)= 2.950×10⁻² mol of NaCl

mole fraction is:

X = 2.950×10⁻² mol of NaCl / (2.950×10⁻² mol of NaCl+55.35mol water) = <em>5.323x10⁻⁴ mol NaCl/Total moles</em>

Mass of NaCl is:

2.950×10⁻² mol of NaCl×(58.44g/mol) = 1.724g of NaCl

Mass of water is:

55.35mol water×(18.02g/mol) = 997.4g of H₂O

(w/w)% is:

1.724g of NaCl / (1.724g of NaCl+997.4g of H₂O) ×100 = <em>0,1726%</em>

<em></em>

Parts per million is mg of NaCl per kg of solution, that is:

1724mg of NaCl / 0.999124g = <em>1726 ppm</em>

<em></em>

I hope it helps!

5 0
3 years ago
From the enthalpies of reaction H2(g)+F2(g)→2HF(g)ΔH=−537kJ C(s)+2F2(g)→CF4(g)ΔH=−680kJ 2C(s)+2H2(g)→C2H4(g)ΔH=+52.3kJ calculate
DIA [1.3K]

Answer:

\Delta H for the given reaction is -2486.3 kJ

Explanation:

The given equation can be written as a combination of the following equation:

2H_{2}(g)+2F_{2}(g)\rightarrow 4HF(g)  ; \Delta H_{1}= (2\times -537kJ)=-1074 kJ

2C(s)+4F_{2}(g)\rightarrow 2CF_{4}(g)  ;  \Delta H_{2}=(2\times -680kJ)=-1360kJ

C_{2}H_{4}(g)\rightarrow 2C(s)+2H_{2}(g)  ;  \Delta H_{3}=-52.3kJ

-----------------------------------------------------------------------------------

C_{2}H_{4}(g)+6F_{2}(g)\rightarrow 2CF_{4}(g)+4HF(g)

\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=(-1074-1360-52.3)kJ=-2486.3kJ

6 0
3 years ago
John walks at a steady speed of 3 mph. How long will it take him to travel 24 miles?
Lynna [10]

Answer:

8 hours i think

Explanation:

4 0
3 years ago
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