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3 years ago

Find the molecular mass of.

Chemistry

2 answers:

dem82 [27]3 years ago

7 0

Answer:

Explanation:

MOLECULAR MASS = ∑ the elements atomic mass from the periodic table

1st NaO

molecular mass = atomic mass of Na + Atomic mass of O

= ( 23 + 16 ) = 39 g/mol

2nd : AlCl3

M.m = atomic mass of Al + Atomic mass of 3Cl

( 27 + 3x35.5 )= 97 g/mole

3rd : Na3(SO4)

M.mass = atomic mass of 3 Na + atomic mass of S + atomic mass of 4 O

( 3x23) + 32 + (4 x 16) =165 g/mol

4th : Mg(OH)2

M.mass = atomic mass of Mg + atomic mass of 2 O + atomic mass of 2

= 24 + 2x16 + 2x1 = 58 g/mol

Margaret [11]3 years ago

3 0

Answer:

sodium oxide Nao
aluminum chloride Al2cl3
sodium sulphate Naso4
magnesium hydroxide Mgoh2

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aksik [14]

Answer:

Explanation:

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3 years ago

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If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit

Svetllana [295]

Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

$4Al + 3O_{2}$ ⇒ $2Al_{2}O_{3}$

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

$16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}$

$10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al$

$19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}$

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

$0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}$

$0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}$

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

Finally, we have to use the yield formula to calculate the percent yield of the reaction:

$Percent of yield = \frac{actual yield}{theoretical yield} * 100 = \frac{0.157 mol Al_{2}O_{3}}{0.1665 mol Al_{2}O_{3}} * 100 = 94.25$

Therefore, the percent of yield is 50.03%.

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3 years ago

What type of substances make better conductors

garri49 [273]

The type of substances that would make good conductors would be substances that has greater density. This would mean that the particles of the substance are much closer to each other and would allow the free flow of the electrons in the substance. With this, current would be able to flow freely.

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4 years ago

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Zn + 2 HCl --> H2 + ZnCl2 If 1.70 g of Zn are reacted, how many grams of ZnCl2 can be created? Show work and process and I wi

Dovator [93]

Explanation:

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