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2 years ago

According to Niels Bohr, electrons travel around an atom's nucleus in well-defined orbits.

Chemistry

2 answers:

Rainbow [258]2 years ago

7 0

Answer:

..........

Explanation:

Bas_tet [7]2 years ago

5 0

Answer: The answer is B

Explanation:

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Explain how the following changes would affect the rate of the reaction of 2-bromo-3-methylbutane with methanol: Part A The alky

swat32

Answer:

The reaction will be slower because the leaving group will be poorer.

Explanation:

One Important thing to put at the back of our mind is that weak bases are good leaving groups. Another thing to take note is that in the halogen series(which is our main subject in this question) as you go down the group the greater or the more heavy the halide is and the heavier halides are more stable that is Bromine will be more stable than chlorine.

Now, we are now told that the bromo halide is been replaced by the chloro halide which means that the rate of Reaction will surely decrease because the leaving group. The cl^- is a stronger base.

6 0

3 years ago

A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant K_f = 5.32 de

kirill115 [55]

Answer: $-15.4^00C$

Explanation:

$T^0_f-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = boiling point of solution = ?

$T^o_f$ = boiling point of solvent (X) = $-10.1^oC$

$k_f$ = freezing point constant = $5.32^oC/kgmol$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte like urea)

= mass of solute (urea) = 29.82 g

= mass of solvent (X) = 500.0 g

$M_2$ = molar mass of solute (urea) = 60 g/mol

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

$(-10.1-(T_f))^oC=1\times (5.32^oC/m)\times \frac{(29.82g)\times 1000}{60\times (500.0g)}$

$T_f=-15.4^0C$

Therefore, the freezing point of solution is $-15.4^0C$

7 0

3 years ago

Copper(II) carbonate undergoes thermal decomposition. One of the pro

Llana [10]

Answer: The green copper (II) carbonate $(CuCO_3)$ changes to black copper oxide $(CuO)$

$CuCO_3(s)\stackrel{\Delta }\rightarrow CaO(s)+CO_2(g)$

Explanation:

Decomposition is defined as the chemical reaction in which a single compound gives two or more simple substances. It requires energy to break the bonds between reactants, thus is an endothermic process.

Thermal decomposition uses heat for decomposition.

The chemical equation for thermal decomposition of copper (II) carbonate is:

$CuCO_3(s)\stackrel{\Delta }\rightarrow CaO(s)+CO_2(g)$

The green copper (II) carbonate $(CuCO_3)$ changes to black copper oxide $(CuO)$

8 0

3 years ago

Why is xenon monatomic but bromine is diatomic

WARRIOR [948]

B 3-4 is the auto
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2 years ago

2.37 x 0.004 Sig Fig

NeTakaya

Answer:

The answer is 0.00948

Explanation:

5 0

3 years ago

Read 2 more answers