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vredina [299]
3 years ago
13

Bohr looked to improve upon Rutherford’s model of the atom because Bohr thought that Rutherford’s model:

Chemistry
1 answer:
AleksandrR [38]3 years ago
8 0
Rutherford's model didn't not account for the properties of electrons. bohr placed electrons in layers of orbit so C
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How do i open the link when i get a question answered​
Feliz [49]

Answer:

Just click on the answer, if it is an external link DO NOT CLICK IT. it may contain dangerous stuff including h*cked stuff.

Explanation:

7 0
3 years ago
A 5 mole sample of liquid acetone is converted to a gas at 75.0°C. If 628 J are required to raise the temperature of the liquid
velikii [3]
The total energy required for this conversion is equivalent to the sum of the energies that are used. There are three steps:

1) Heating of liquid acetone
This used 628 J

2) Evaporation of acetone
This used 15.6 kJ or 15,600 J

3) Heating of acetone vapors
This used 712 J

Adding these quantities,

Total energy = 628 + 15,600 + 712

The total energy required was <span>16940 Joules of 16.94 kJ</span>
7 0
3 years ago
Na + CI2 =2NaC is It balanced
scoray [572]

Answer:

no, the correct answer is NaCI

Explanation:

you're welcome

3 0
3 years ago
What is the formula of the chromium(iii) complex that contains two ammonia and four thiocyanate (scn−) ligands?
Nonamiya [84]
Answer is: formula of the complex is [Cr(NH₃)₂(SCN)₄<span>]</span>⁻<span>.
This complex has negative charge (-1) because chromium (central atom or metal) has oxidation number +3, first ligand ammonia has neutral charge and second ligand thiocyanate has negative oxidation number -1:
+3 + 2</span>·0 + 4·(-1) = -1.
4 0
3 years ago
sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction
prisoha [69]

Answer:

2.76 × 10⁻¹¹  

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

                         C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹:       0         0         -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

\text{The relationship between $\Delta G^{\circ}$ and K  is}\\\Delta G^{\circ} = -RT \ln K

T = (25.0 + 273.15) K = 298.15 K

\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}

3 0
3 years ago
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