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Arada [10]
3 years ago
6

The molal boiling-point and freezing-point constants are different for different solvents.

Chemistry
2 answers:
RoseWind [281]3 years ago
8 0

Answer:

I think the answer is true, sorry if I am wrong

Explanation:

Vadim26 [7]3 years ago
3 0
The right answer for this is false
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Gamma rays

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in rutherford's gold foil experiment a very small number of alpha particles were deflected. what about the structure of the atom
zvonat [6]

Answer:The atom being mostly empty space. A small number of alpha particles were deflected by large angles (> 4°) as they passed through the foil. There is a concentration of positive charge in the atom. Like charges repel, so the positive alpha particles were being repelled by positive charge

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3 years ago
Explain why 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane. hint: examine the
Mama L [17]

Answer:-

The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.

The I - attacks from backside to give the transition state for both.

If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.

This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.

7 0
3 years ago
Los enlaces covalentes se presentan tanto en sólidos moleculares como en los que forman una red covalente ¿porque estas clases d
Viefleur [7K]

Answer:can u anser my question

Explanation: On page 17, author Joey Bartolomeo writes “Caleb participated in the investigation by testifying about what happened during the health seminar at his school.” As it is used in that sentence, what does "testify" mean? *

0 points

C. to persuade someone to join a group

B. to disguise or hide from sight

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3 0
2 years ago
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

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3 years ago
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