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mario62 [17]
3 years ago
12

if an element A has an ionic charge of +2 and element B has an ionic charge of -3, what will be the formula for the compound cre

ated by element A and element B?
Chemistry
1 answer:
KengaRu [80]3 years ago
8 0

If element A has an ionic charge of +2 and element B has an ionic charge of -3

then the reaction is given as:  A^2+ + B^3- --------------------> A3B2

the compound formed is : A3B<span>2</span>

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You weigh a 24K gold chain and find that it weighs 400 grams. Determine home many moles of gold atoms you have
makkiz [27]

Answer:

                     2.03 moles of Gold

Explanation:

                     Gold is one of the most precious metal metal used in many applications and mainly as a jewellery. In terms of purity it is categorized in Karats. 24 Karat is considered the purest Gold (i.e. 100 % Gold) while other Karats (14, 18, 22 e.t.c) are alloys with other metals and gyms.

Data Given:

                 Mass of Gold  =  400 g

                 A.Mass of Gold  =  196.97 g.mol⁻¹

Calculate Moles of Gold as,

                              Moles  =  Mass ÷ M.Mass

Putting values,

                              Moles  =  400 g ÷ 196.97 g.mol⁻¹

                              Moles  =  2.03 moles of Gold

4 0
2 years ago
How many control(s) are in an experiment
weqwewe [10]
You can have as many controls as necessary, But they must remain equal at all times in order to get the most accurate results
6 0
3 years ago
If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

3 0
2 years ago
Calculate the volume of water which was liberated during the decomposition of hydrogen peroxide.
nikitadnepr [17]

The answer is :

The normal equation for the decomposition of hydrogen peroxide indicates that 2 vol. peroxide vapor should give rise to 2 vol. water vapor and 1 vol. oxygen.

What is hydrogen peroxide ?

  • Hydrogen peroxide is water (H2O) with an extra oxygen molecule (H2O2).
  • The extra oxygen molecule oxidizes, which is how peroxide gets its power says Dr. Beers.
  • This oxidation kills germs and bleaches color from porous surfaces like fabrics.
  • Hydrogen peroxide is found in biological systems including the human body.
  • Enzymes that use or decompose hydrogen peroxide are classified as peroxidases.

To learn more about Hydrogen peroxide visit: brainly.com/question/18709693?

#SPJ4

4 0
1 year ago
If 4.0 g of hydrogen and 10.0 g of oxygen are mixed according to the equation 2 h2 + o2 → 2 h2o, which is the limiting reagent?
mars1129 [50]
From the equation, we can see that the molar ratio between hydrogen and oxygen is:
2 : 1

Next, we determine the moles of hydrogen and oxygen that are actually present using:
moles = mass / Mr

Hydrogen:
moles = 4 / 2 = 2

Oxygen:
10/32 = 0.3125

Therefore, it is evident that the moles of oxygen present, 0.3125, are less than those that are required for 2 moles of hydrogen, which is 1. This makes oxygen the limiting reactant, which is the one that limits the completion of a reaction.
8 0
3 years ago
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