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3 years ago

Al2(SO3)3 what are the number of atoms

1 answer:

4 0

Answer:In the chemical formula Al2(SO4)3, the Al2 means there two aluminium (atoms or ions). The SO4 is a sulfate ion and (SO4)3 means there are 3 sulfate ions. The number 3 before Al2(SO4)3 means there are three times the number of atoms and ions of the chemical formula.

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For each reaction below, use the drop-down menus to select which compound will form a precipitate. Note: It is possible that no

Dimas [21]

Answer:

1. Cadmium sulfide

2. Iron (III) hrydroxide

3. Barium carbonate

8 0

2 years ago

A sample of oxide iron weighing 2.40g was heated in a stream of hydrogen until it was completely converted to the metal. If the

wlad13 [49]

Answer:

Fe₂O₃

Explanation:

To solve this question we must find the moles of Iron in 1.68g. With the difference of the masses we can find the moles of oxygen. The formula will be obtained with the ratio of both amount of moles:

<em>Moles Fe:</em>

1.68g * (1mol / 56g) =0.03moles

<em>Moles O:</em>

2.40g-1.68g = 0.72g * (1mol/16g) = 0.045moles

The ratio O/Fe is:

0.045moles / 0.03moles = 1.5 moles. this ratio is obtained if the formula is:

4 0

3 years ago

If an element has two isotopes, what is the atomic mass if one of the isotopes has a mass of 15.000 amu and makes up 5.000% of t

olasank [31]

The atomic mass of element is the weighted average atomic mass of the element with respect to the abundance of the isotopes of that element
atomic mass is the sum of the products of the mass of isotopes by their percentage abundance
atomic mass = 15.000 amu x 5.000 % + 16.000 amu x 95.000 %
= 0.7500 + 15.200
atomic mass of element is therefore 15.950

8 0

3 years ago

3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)

GrogVix [38]

Explanation:

(a) The given reaction equation is as follows.

$Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g)$ (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$

Reduction-half-reaction: $NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)$

As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

$Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l)$ (acidic)

(b) The given reaction equation is as follows.

$HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq)$ (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: $HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)$

Oxidation-half reaction: $Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}$

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

$4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}$

Therefore, balancing the whole reaction equation in the basic medium as follows.

$H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)$

6 0

3 years ago

Given a 0.5 M stock solution of aqueous NaCl, how much of the stock solution would you need to make a 50 mL sample of 0.25 M aqu

MrRissso [65]

Answer:

25 ml of stock

Explanation:

Molarity x Volume of Concentrated Stock = Molarity x Volume of Diluted Stock

=> Mc x Vc = Md x Vd => (0.5M)(x) = (0.25M)(50 ml)

=> x = Vc = (0.25M x 50 ml) / (0.50M) = 25 ml of stock

7 0

3 years ago