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3 years ago

7. This part of the brain controls memory, and long term abuse of marijuana can permanently

Chemistry

2 answers:

Hatshy [7]3 years ago

8 0

The answer is the third one which is Hippocampus

eduard3 years ago

7 0

Answer:

Hippocampus

Explanation:

The main parts of the brain involved with memory are the amygdala, the hippocampus, the cerebellum, and the prefrontal cortex.

Memory impairment from marijuana use occurs because THC alters how the hippocampus, a brain area responsible for memory formation, processes information.

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Describe how elements are laid out on the periodic table of elements.

defon

Answer:

The chemical elements are arranged in order of increasing atomic number.

Explanation:

Hydrogen = 1

Helium = 2

Lithium = 3

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3 years ago

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3 years ago

Enter the net ionic equation, including phases, for the reaction of AgNO3 (aq) and KCl(aq). Refer to the solubility rules as nee

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4 years ago

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A 20.7167 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation mgco3(s) → mgo(s

marishachu [46]

The mass of impure $MgCO_{3}$ is 20.7167 g. The decomposition reaction is as follows:

$MgCO_{3}(s)\rightarrow MgO(s)+CO_{2}(g)$

Here, 1 mole of $MgCO_{3}$ gives 1 mole of MgO.

Molar mass of $MgCO_{3}$ and MgO is 84.31 g/mol and 40.3044 g/mol respectively.

Converting number of moles in terms of mass,

$n=\frac{m}{M}$

Here, M is molar mass.

Since, $n_{MgCO_{3}}=n_{MgO}$

Thus, $\frac{m_{MgCO_{3}}}{M_{MgCO_{3}}}=\frac{m_{MgO}}{M_{MgO}}$

On putting the values,

$\frac{m_{MgCO_{3}}}{84.31}=\frac{m_{MgO}}{40.3044}$

Rearranging,

$m_{MgO}=\frac{m_{MgCO_{3}}}{84.31}\times 40.3044$

Or,

$m_{MgCO_{3}}=\frac{m_{MgO}}{0.4780}$ ...... (1)

Let the mass of impurity be $M_{I}$ and mass of impure $MgCO_{3}$ is 20.7167 g thus,

$M_{MgCO_{3}}=(20.7167-M_{I})g$ ...... (2)

Also, mass of impure MgO is 16.8817 g thus,

$M_{MgO}=(16.8817-M_{I})g$ ...... (3)

On comparing equations (2) and (3),

$M_{MgO}=M_{MgCO_{3}}-3.835$

Putting the value of $M_{MgO}$ in equation (1),

$m_{MgCO_{3}}=\frac{M_{MgCO_{3}}-3.835}{0.4780}$

Or,

$0.4780 M_{MgCO_{3}}=M_{MgCO_{3}}-3.835$

Or,

$M_{MgCO_{3}}-0.4780M_{MgCO_{3}}=3.835$

Or,

$0.522 M_{MgCO_{3}}=3.835$

Or,

$M_{MgCO_{3}}=\frac{3.835}{0.522}=7.35 g$

Thus, magnesium carbonate present in the original sample is 7.35 g.

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3 years ago

Which compound listed below has the largest molar mass? Ca(OH)2 H20 CaCl2 NH3

Crazy boy [7]

Answer: the answer would be CaCI2

Explanation: this was one of my exam questions i got right. :)

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3 years ago