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Crank
2 years ago
15

Pls help, needed for chemistry

Chemistry
1 answer:
vovangra [49]2 years ago
7 0

Answer:

go to a calculator and see the answer then make the hypothisis which is the answer

Explanation:

magic

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How many significant figures are in 195.01
Masja [62]

Answer:

difficult to answer, need a bit more detail

5 0
3 years ago
What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?
Vsevolod [243]

The dilution formula can be used to find the volume needed

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

c1 - 0.33 M

c2 - 0.025 M

v2 - 25 mL

Substituting these values in the equation

0.33 M x v1 = 0.025 M x 25 mL

v1 = 1.89 mL

Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution

7 0
3 years ago
. Consider the following half-reactions:
Vedmedyk [2.9K]

d. Fe(s) and Al(s)

<h3>Further explanation</h3>

In the redox reaction, it is also known  

Reducing agents are substances that experience oxidation  

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series  

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

The electrodes which are easier to reduce than hydrogen (H), have E cells = +

The electrodes which are easier to oxidize than hydrogen have a sign E cell = -

So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)

The metal  : d. Fe(s) and Al(s)

7 0
3 years ago
Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.
Dovator [93]
Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10  5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2  2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0
3 years ago
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What are 2 diseases caused by coronaviruses?
ivann1987 [24]

Answer:

SARS and MERS

Explanation:

8 0
3 years ago
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