Why is magnification important to the study of life science?

Chemistry

1 answer:

trapecia [35]3 years ago

5 0

Answer:

magnification makes everything smaller so you can see smaller things up close and study them more

Explanation:

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At 1.75 atm, a sample of gas takes up 2.88L. If the pressure in the gas is increased to 2.25 atm, what will the new volume be

yarga [219]

Explanation:

P1V1= P2V2

V2= P1V1/P2

= 1.75*2.88/2.25 = 2.24

5 0

3 years ago

A solution of sucrose in water is 48.5% sucrose by mass and has a density of 1.118 g/ml. what mass of sucrose, in grams, is cont

telo118 [61]

To determine the mass of sucrose from a given volume of solution, we need to convert the volume into mass by using the density of the solution. We calculate as follows:

mass solution = 3.50 ( 1118 ) = 3913 g

mass of sucrose = 3913 g solution ( .485 g sucrose / g solution ) = 1897.805 g sucrose is present in the solution.

6 0

3 years ago

Read 2 more answers

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$