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FromTheMoon [43]
3 years ago
10

Which model of the atom is most accurate?

Chemistry
1 answer:
cricket20 [7]3 years ago
3 0

Answer:

it's the electron cloud model

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What mass of slaked lime is needed to decompose 10 g of ammonium chloride to give 100% yield? Ca(OH)2 + 2NH4Cl → CaCl2 + 2NH3 +
dimulka [17.4K]

Answer is: mass of slaked lime is 6.92 grams.

Balanced chemical reaction: Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O.

m(NH₄Cl) = 10 g; mass of ammonium chloride.

M(NH₄Cl) = 14 + 1·4 + 35.5 · g/mol.

M(NH₄Cl) = 53.5 g/mol; molar mass of ammonium chloride.

n(NH₄Cl) = m(NH₄Cl) ÷M(NH₄Cl).

n(NH₄Cl) = 10 g ÷ 53.5 g/mol.

n(NH₄Cl) = 0.187 mol; amount of ammonium chloride.

From balanced chemical reaction: n(NH₄Cl) : n(Ca(OH)₂) = 2 : 1.

n(Ca(OH)₂) = 0.093 mol.

m(Ca(OH)₂) = n(Ca(OH)₂) · M(Ca(OH)₂).

m(Ca(OH)₂) = 0.093 mol · 74.1 g/mol.

m(Ca(OH)₂) = 6.92 g.

8 0
3 years ago
What is the ΔG given the following? ΔH = 20 kJ/mol T = 15°C ΔS = .101 kJ/molK
DedPeter [7]
∆G = ∆H-T∆S
=20×10^3-(15+273)(.101×10^3)
=20000-(288)(101)
=20000-29088
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6 0
3 years ago
Which statement describes one feature of Rutherford model of the atom
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\huge{ \color{magenta}{ \fcolorbox{magenta}{black}{ \huge{ \color{white}{ \fcolorbox{aqua}{black}{♡Answer♡ }}}}}}

<em><u>The Rutherford model shows that an atom is mostly empty space, with electrons orbiting a fixed, positively charged nucleus in set, predictable paths.</u></em>

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2 years ago
Jordan needs to repeat the experiment but his teacher says that he needs to improve his design in his second experiment what sho
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Just add more detail in the second experiment explain every little thing.
8 0
3 years ago
Read 2 more answers
A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.
Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

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

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K

Putting values in above equation, we get:

\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L

Therefore, the new volume of the air is, 6.83 L

6 0
3 years ago
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