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3 years ago

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write the chemical formula for the following compound -magnesium chloride, zinc sulphide Sodium,sodium sulphate, calcium bicarbo

nate.

1 answer:

Sindrei [870]3 years ago

6 0

Answer:

magnesium chloride: MgCl2

zinc sulphide: ZnS

sodium sulphate: Na2SO4

calcium bicarbonate: Ca(HCO3)2

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Which answer lists the fundamental forces in order from strongest to weakest

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Actually, gravity is the weakest of the four fundamental forces. Ordered from strongest to weakest, the forces are 1) the strong nuclear force, 2) the electromagnetic force, 3) the weak nuclear force, and 4) gravity.May 22, 2013

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Who invented peroidic table

ch4aika [34]

<span>Dmitri Mendeleev thats the answer</span>

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If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for

Luba_88 [7]

The number of grams of Ag2SO4 that could be formed is 31.8 grams

<u><em> calculation</em></u>

Balanced equation is as below

2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)

Find the moles of each reactant by use of mole= mass/molar mass formula

that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles

moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles

use the mole ratio to determine the moles of Ag2SO4

that is;

the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles

The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles

AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles

<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>

that is 0.102 moles x 311.8 g/mol= 31.8 grams

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3 years ago

A solution contains 6.21 g of ethylene glycol dissolved in 25.0 g of water. If water has a boiling point elevation constant of 0

timurjin [86]

Calculate first the number of moles of ethylene glycol by dividing the mass by the molar mass.
n = (6.21 g ethylene glycol) / 62.1 g/mol
n = 0.1 mol
Then, calculate the molality by dividing the number of moles by the mass of water (in kg).
m = 0.1 mol/ (0.025 kg) = 4m
Then, use the equation,
Tb,f = Tb,i + (kb)(m)
Substituting the known values,
Tb,f = 100°C + (0.512°C.kg/mol)(4 mol/kg)
<em>Tb,f = 102.048°C</em>

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3 years ago

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what is the empirical formula of vanadium 1 oxide given that 20.38 grams of vandium combines with oxygen to form 23.58 grams of

Alex

Answer:

The empirical formula is V₂O

Empirical formula of a compound is the formula that shows the simplest whole number ratio of the atoms of elements in a given compound. Empirical formula is normally calculated when the mass of each element in a compound is known or the percentage composition by mass of each element in a compound is known.

Step by Step Explanation:

Step 1: Percentage composition of each element

Percentage composition=(mass of an element/ mass of the compound)100%

Mass of Vanadium = 20.38 g

Mass of the compound = 23.58 g

% composition of Vanadium = (20.38 g/23.58 g) 100%

= 86.43 %

Mass of Oxygen = 23.58 g -20.38 g

= 3.2 g

% composition of oxygen = (3.2/g/23.58 g) 100%

= 13.57%

Step 2: Find the number of atoms of each element in the compound

Number of atoms = percentage composition/ atomic mass

Atomic mass of Vanadium = 50.94 g/mol

Number of atoms of V = 86.43 /50.94

= 1.6967

Atomic mass of oxygen = 16 g/mol

Number of atoms of O = 13.57/16

= 0.8481

Step 3: Find the simplest ratio of atoms

Vanadium : Oxygen

1.6967 : 0.8481

= 1.6967/0.8481 : 0.8481/0.8481

= 2: 1

Whole number ratio = 2 : 1

Therefore; the empirical formula is V₂O

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3 years ago

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