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sp2606 [1]
3 years ago
13

Plz help!!!

Chemistry
1 answer:
baherus [9]3 years ago
7 0

a) The change here is that metallic iron is converted into ions and copper is deposited. This is called a displacement reaction.

b) \text{Fe} is oxidised in this reaction.

c)$ \mathrm{Fe_{(s)}+ CuSO_{4(aq)} \rightarrow FeSO_{4(aq)} + Cu_{(s)}}$

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La masa de una olla es de 300g y contiene 90% de aluminio. Hallar el número de moles de aluminio de la olla. P.A.(Al= 27)
Neko [114]

Explanation:

The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)

The mass of aluminum present in the pot is:

300 g * 90/100\\=270 g

Hence, in the given pot 270g Al is present.

Number of moles of Al=\frac{given mass ofAl}{its molar mass}

The gram atomic mass of Al -27 g/mol

Given the mass of Al is 270 g

Substitute these values in the above formula:

Number of moles of Al=\frac{given mass ofAl}{its molar mass}\\=\frac{270 g}{27 g} \\=10.0 mol

Answer is 10.0 mol of Al is present.

6 0
3 years ago
Precipitation _____. occurs equally all over the globe. is the change in state from a gas to a liquid. happens when ice changes
babunello [35]

Precipitation exceeds evaporation over the land.

5 0
3 years ago
How many grams of pure naoh must be used to prepare 10.0 l of a solution that has a ph of 13? __________g?
padilas [110]

40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.

<em>Step 1</em>. Calculate the pOH of the solution

pOH = 14.00 – pH = 14.00 -13 = 1

<em>Step 2</em>. Calculate the concentration of NaOH

[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH

<em>Step 4</em>. Calculate the mass of NaOH

Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH

8 0
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What is the final concentration of a solution prepared by diluting 35.0 ml of 12.0 m hcl to a final volume of 1.20 l? what is th
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We know that, M1V1     =     M2V2
                        (Initial)          (Final)
where, M1 and M2 are initial and final concentration of soution respectively. 
V1 and V2 = initial and final volume of solution respectively

Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml

∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
8 0
3 years ago
What is the molar mass of Fe2Br3?
aleksklad [387]
<span>351.4020 g/mol which is 3.51 x 10</span>∧ 2 g/mole
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