True or False? You should change the subscripts to balance chemical equations.

What's the definition of ions

Masja [62]

An atom or molecule with a net electric charge due to the loss or gain of one or more electrons. :)

3 0

3 years ago

Read 2 more answers

How many liters of solution can be produced from 2.5 moles of solute if a 2.0 M

Firlakuza [10]

Answer:

The answer is 1.25 L

Explanation:

5 0

2 years ago

William adds two values , following the rules for using significant figures in computations. He should write the sum of these tw

yuradex [85]

Answer:

when it comes to adding or subtracting numbers, his final answer should have the same number of decimal places as the least precise value.

For example if you add 2 numbers; 10.443 + 3.5 , 10.443 has 3 decimal places and 3.5 has only one decimal place.

Therefore 3.5 is the less precise value.

So when adding these 2 values the final answer should have only one decimal place.

after adding we get 13.943 but it can have upto one decimal place. then the second decimal place is less than 5 so the answer should be rounded off to 13.9.

the answer is the same number of decimal places as the least precise value

Explanation:

I think this is the answer I'm not sure

7 0

2 years ago

15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula

otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C = 15.89 % Carbon

21.18 g of O = 21.18 % Oxygen

62.93 g of Os = 62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 15.89 g/ 12 g/mol

no. of mole = 1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 21.18 g/ 16 g/mol

no. of mole =

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 62.93 g/ 190 g/mol

no. of mole =

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

7 0

3 years ago

4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.

slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ $\frac{1mol}{102g}$ = 0,0980 moles

And 10,0g of HCl are:

10,0 gₓ $\frac{1mol}{36,5g}$ = 0,274 moles

For a total reaction of 0,274 moles of HCl you need:

0,274× $\frac{1molesAl_{2}O_3}{6 mole HCl}$ = 0,0457 moles of Al₂O₃

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl × $\frac{2 moles AlCl_{3}}{6 moles HCl}$ × 133 $\frac{g}{mol}$ = 12,1 g of AlCl₃

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles = 0,0523 moles

And its mass is:

0,0523 molesₓ $\frac{102g}{1mol}$ = 5,33 g of Al₂O₃

I hope it helps!

7 0

3 years ago