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g100num [7]
3 years ago
11

SOMEONE HELP ASAPP...

Chemistry
2 answers:
saveliy_v [14]3 years ago
4 0

AnswU GOT THIS!!!!!!!!!!!!!

Explanation:

Alisiya [41]3 years ago
3 0

Answer:

protein??

Explanation:

im not sure, but I could be wrong!!

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A chemical reaction in which one element replaces another element in a compound can be categorized as a
rewona [7]
They are substitution reactions 

hope that helps
7 0
3 years ago
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Which of the following is NOT a chemical reaction?
Damm [24]
Ice melting from water. Nothing has changed except the state of matter.
6 0
3 years ago
Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat
Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

  • Density of aluminum (ρ): 2.7 g/cm³
  • Mass of aluminum (m): 81 g
  • Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

4 0
3 years ago
3. What noble gas would be part of the electron configuration notation for Mn?
shusha [124]

Answer:

Argon {Ar}

Explanation:

The noble gas used for a condensed electron configuration is the one before the element which you are configuring.

In this case, the element (Mn) is manganese

The noble gas that is before this element is Argon which is the row above it

so your configuration would be {Ar} 3d^5 4s^2

7 0
3 years ago
Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?
Andrews [41]
Answer: 
 Zn =⇒ Zn+2(0.10) + 2e- (anode)
 Zn+2(?M) + 2e- === Zn(s) (cathode)
 
 Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn 
 E = E^o -0.0592 log Q; in this case E^o is zero. 
 E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2 
 23 mV x 1 volt/1000mv = 0.023 Volts 
 0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
 0.023 = -0.0296 { log 0.10 – log [Zn+2] }
 0.023 = -0.0296{ -1 - log[Zn+2] }
 0.023 = +0.0296 + 0.0296log[Zn+2]
 -0.0066 = 0.0296log[Zn+2]
 -0.22= log[Zn+2]
 [Zn+2] = 10^-0.22 = 0.603 Molar

6 0
3 years ago
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