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3 years ago

Al2O3 would be typed in as Al2O3 Al2(SO4)3 would be typed in as Al2(SO4)3.

Chemistry

1 answer:

N76 [4]3 years ago

4 0

Answer:

Explanation:

Hello,

We are required to write the chemical formula of the following compounds

1. Calcium nitride = Ca₃N₂

2. Magnesium Phosphide = Mg₃P₂

3. Rubidium Chromate = Rb₂CrO₄

4. Aluminium nitrate = Al(NO₃)₃

5. Ammonium Arsenide = (NH₄)₃As

6. Nickel(ii) nitrite = Ni(NO₂)₂

7. Copper(i)sulfate = Cu₂SO₄

8. Iron(iii)nitrate = Fe(NO₃)₃

9. Manganese(ii)nitrate = Mn(NO₃)₂

The left hand side are the common names when the right hand side are the chemical formula.

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What is ancient japan remembered for

Katarina [22]

Hi!

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3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

2 years ago

When preparing the diazonium salt, the solution is tested with potassium iodide-starch paper. a positive test is the immediate f

Dahasolnce [82]

KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which <span>then will react with 4-nitroaniline to form a diazonium salt.</span>

8 0

2 years ago

How many atoms are in 3 NaCl?

Assoli18 [71]

Explanation:

Sodium Chloride or NaCl is made up of two elements, sodium (or Na) and chlorine (or Cl). A molecule of sodium chloride, NaCl, consists of one atom each of sodium and chlorine. Hence, each molecule of NaCl has 2 atoms total.

5 0

2 years ago

A pure copper penny contains approximately 2.9×1022 copper atoms. Use the following definitions to determine how many ______ of

hammer [34]

Complete question is;

A pure copper penny contains approximately 2.9 × 10^(22) copper atoms.

1 doz = 12

1 gross = 144

1 ream = 500

1 mol = 6.022 × 10^(23)

Use these definitions to determine the following:

A) How many dozens of copper atoms are in a penny.

B) How many gross of copper atoms are in a penny

C) How many reams of copper atoms are in a penny.

D) how many moles of copper atoms are in a penny?

All answers can be rounded to two significant figures

Answer:

A) 2.4 × 10^(21) dozens

B) 2.01 × 10^(20) gross

C) 5.8 × 10^(19) reams

D) 0.048 mol

Explanation:

A) A dozen contains 12.

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/12 dozens = 2.42 × 10^(21).

In 2 significant figures, we have;

2.4 × 10^(21) dozens

B) 1 gross = 144

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/144 gross ≈ 2.01 × 10^(20) gross

C) 1 ream = 500

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/500 reams = 5.8 × 10^(19) reams

D) 1 mol = 6.022 × 10^(23)

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/(6.022 × 10^(23)) = 0.048 mol

6 0

3 years ago