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Margarita [4]
3 years ago
9

When determining the formula for a compound, it is important to know the valence numbers of each element. Suppose you want to wr

ite a formula for a compound with these two elements: Element A = valence 3 Element B = valence 2 In the formula, what number will be the subscript for Element B?
Chemistry
1 answer:
Damm [24]3 years ago
3 0

Explanation:

The valency of the element is the measure of the combining power of the element with the other atoms when the element forms compounds or molecules.

<u>Thus, valency has to known to find the how the elements have been combined and how many electrons have been lost, gain or shared.</u>

Given that:

Element A has 3 valence electrons and element B has 2 valence electrons. To find the ionic compound, the valency of the cations and the anions are interchanged and are written in subscripts. Thus,

A        B

3          2

Cross multiplying, we get the formula : A_2B_3

<u>Hence, 3 is the subscript for Element B.</u>

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A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
IRINA_888 [86]

<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

<h3><em>Learn more </em></h3>

density acetone

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Keywords: density, mass, volume, a gold bar

3 0
3 years ago
Read 2 more answers
calculate the mass of ammonium sulfate that can be made from 51 grams of ammonia. 2NH₃ + H₂SO₄ → (NH₄)₂SO₄ (keeping the mole con
kap26 [50]

Explanation:

firstly firstly we are to calculate the number of moles of ammonia and using the mole concept of two moles of ammonia gives one mole of ammonium sulphate we can calculate the number of moles of ammonium sulphate and mass

from n=m/mr

3 0
2 years ago
Is H2O polar or non polar?
otez555 [7]

Answer:

Explanation:

H2O is a polar molecule

8 0
2 years ago
Read 2 more answers
4 FeCO3 + O2 --&gt; 2 Fe2O3 + 4CO2
Andre45 [30]

The percentage yield of the reaction : 93.3%

Mass of FeCO₃ 2310.44 g

<h3>Further explanation </h3>

The reaction equation is the chemical formula of reagents and product substances  

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products  

Reaction

4 FeCO₃ + O₂ ⇒ 2 Fe₂O₃ + 4CO₂

MW FeCO₃ : 115,854 g/mol

MW Fe₂O₃ : 159,69 g/mol

  • mol FeCO₃

\tt \dfrac{35}{115.854}=0.302

  • mol Fe₂O₃

\tt \dfrac{2}{4}\times 0.302=0.151

  • mass Fe₂O₃

\tt 0.151\times 159.69=24.113~g

  • percent yield

\tt \dfrac{22.5}{24.113}\times 100\%=93.3\%

mol of 1 kg Fe₂O₃ = 1000 g

\tt \dfrac{1000}{159.69}=6.262

mol of FeCO₃

\tt \dfrac{4}{2}\times 6.262=12.524

mass of FeCO₃

\tt 12.524\times 115.854=1450.955~g

a purity of 62.8%

\tt \dfrac{100}{62.8} \times 1450.955=2310.44~g

4 0
3 years ago
Please help me?!!!!!!!
SCORPION-xisa [38]

Answer:

this is true! good luck!

4 0
3 years ago
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