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mel-nik [20]
3 years ago
15

How many moles are equal to 1.3 x 10^24 atoms of aluminum?

Chemistry
1 answer:
emmasim [6.3K]3 years ago
7 0

Answer:

Explanation: well I think 6.02 x 10^23 is 1 mole. So do 1/6.02 x 10^23 = x/1.3 x 10^24

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how do you know the state of a compound when writing out the equation. WILL MARK BRAINLIEST PLEASE HELP!!
tester [92]

Answer:

The state of matter of each compound or molecule is indicated in subscript next to the compound by an abbreviation in parentheses. For example, a compound in the gas state would be indicated by (g), solid (s), liquid (l), and aqueous (aq).

Explanation:

3 0
3 years ago
I need this answer quick please show work
Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

4 0
3 years ago
Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

4 0
3 years ago
WHAT BLOCK IN THE PICTURE HAS:
GREYUIT [131]

Answer:

the highest is "A" and the lower is "C"

7 0
3 years ago
Sjwjdbehdhenwhdbrjxbejsnejfbejdnfbrjrnd
Keith_Richards [23]

Answegood

Explanation:this answer is two it has letters

6 0
2 years ago
Read 2 more answers
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