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dedylja [7]
3 years ago
10

A patient comes in showing signs of severe depression. Dr. Jorgenson wants to order an EEG and hormone panel, while Dr. Frieden

wants to set up a system of rewards and punishments, demonstrating the conflicting approaches of
Chemistry
1 answer:
Citrus2011 [14]3 years ago
6 0
So what is the question asking exactly?
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Separate this redox reaction into its balanced component half-reactions. What is the oxidation and reduction half reactions
erma4kov [3.2K]
<span>Separate this redox reaction into its component half-reactions. 
Cl2 + 2Na ----> 2NaCl 

reduction: Cl2 + 2 e- ----> 2Cl-1 
oxidation: 2Na ----> 2Na+ & 2 e- 


2) Write a balanced overall reaction from these unbalanced half-reactions: 

oxidation: Sn ----> Sn^2+ & 2 e- 
reduction: 2Ag^+ & 2e- ----> 2Ag 

giving us 
2Ag^+ & Sn ----> Sn^2+ & 2Ag </span>Steve O <span>· 5 years ago </span><span>
</span>
3 0
3 years ago
Read 2 more answers
The area that includes everything between the cell membrane and the nucleus of a cell is called the
kolbaska11 [484]

Answer:

Cytoplasm.

Explanation:

Please mark as Brainliest! :)

8 0
3 years ago
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If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc
hodyreva [135]

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

Mass of cryolite (Na_3AlF_6) = moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg

Thus 69.72 kg of cryolite will be produced.

8 0
3 years ago
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
Read 2 more answers
An astronaut weighs 104 newtons on the moon,where the strength of gravity is 1.6 newtons per kilogram
belka [17]
I'd say he ways about 35 kilograms, but I'm probably wrong, xD

6 0
3 years ago
Read 2 more answers
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