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adell [148]
3 years ago
7

Which of the following is not an example of a base? KOH LiOH HCl NaOH

Chemistry
1 answer:
Inessa05 [86]3 years ago
7 0
HCI, so C HCI is an acid
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What are natural resources
Angelina_Jolie [31]
Resources that come from nature, some examples include: Water, Gold, Oil, Coal, Apples, Oranges, etc. 
5 0
3 years ago
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Iron is a solid, grey metal. Oxygen is a colorless gas. When iron and oxygen chemically combine , rust is made. Rust has a reddi
Phoenix [80]
Because the formation of rust is a kind of chemistry reaction.
After chemistry reaction, compound with new properties is produced.
Of course, color is one of the new properties.
For example, if you put iron in colorless acid solution, a green solution is made.
This kind of chemisty reaction differs from physical reaction, for example, pigment with one color dissolves in liquid, a liquid with same color is made.
6 0
3 years ago
Using the van der waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 c is____________at
MatroZZZ [7]

Answer:

D. 1.48atm

Explanation:

Van der waals equation is given as:

(P +an²/v²) (v - nb) = nRT

Where;

P = pressure (atm)

V = volume (L)

R = gas constant (0.0821 Latm/molK)

a and b = gas constant specific to each gas

T = temperature (K)

n = number of moles

According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol

Hence;

(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15

(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638

(P + 0.0291) (22.316) = 33.638

22.316P + 0.649 = 33.638

22.316P = 33.638 - 0.649

22.316P = 32.989

P = 32.989/22.316

P = 1.478

P = 1.48atm

6 0
3 years ago
The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell
tankabanditka [31]

Explanation:

In a voltaic cell, oxidation reaction occurs at anode whereas reduction reaction occurs at the cathode.

Hence, the half-cell reaction taking place at anode and cathode will be as follows.

At anode (Oxidation) : Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-} ...... (1)

At cathode (Reduction) : Ag^{+}(aq) + 1e^{-} \rightarrow Ag(s)

So, in order to balance the half cell reactions, we multiply reduction reaction by 3. Hence, reduction reaction equation will be as follows.

              3Ag^{+}(aq) + 3e^{-} \rightarrow 3Ag(s)  ........ (2)

Therefore, overall reaction will be sum of equations as (1) + (2). Thus, net reaction equation is as follows.

      Cr(s) 3Ag^{+}(aq) \rightarrow Cr^{3+}(aq) + 3Ag(s)          

         

6 0
3 years ago
I don't know how to do this can I get the answers plz it's due in 1 hour
katrin [286]
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.
4 0
3 years ago
Read 2 more answers
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