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3 years ago

Determine whether or not each chemical formula is an empirical formula. C6H12O6 CaCO3 NaMnO4 Ba3(PO4)2 K2C2O4

Chemistry

1 answer:

zavuch27 [327]3 years ago

5 0

Answer:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Explanation:

Hello!

In this case, since the empirical formulas are referred to those formulas that have been simplified up to the smallest whole number for each subscript per atom, we can see that:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Best regards!

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Which of these is an isoelectronic series? 1) na+, k+, rb+, cs+ 2) k+, ca2+, or, s2– 3) na+, mg2+, s2–, cl– 4) li, be, b, c 5) n

ss7ja [257]

An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:

Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6

A. Na⁺: 11-1 = 10 electrons
K⁺: 19 - 1 = 18 electrons
Rb⁺: 37-1 = 36 electrons

B. K⁺: 19 - 1 = 18 electrons
Ca²⁺: 20 - 2 = 18 electrons
S²⁻: 16 +2 = 18 electrons

C. Na⁺: 11-1 = 10 electrons
Mg²⁺: 12 - 2 = 10 electrons
S²⁻: 16 +2 = 18 electrons

D. Li=3 electrons
Be=4 electrons
B=5 electrons
C=6 electrons

The answer is letter B.

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3 years ago

A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to

melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

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A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.

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The question is incomplete, complete question is:

A fluid occupying $3.2 m^3$ of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = $3.2 m^3$

Density of the fluid = D

$D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3$

Specific volume is the reciprocal of the density :

$V_{specific}=\frac{1}{Density}$

Specific volume of the fluid = $S_v$

$S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg$

Density of the fluid in English Engineering units = D $(lb/ft^3)$

1 kg = 2.20462 lb

1 m = 3.280 ft

$D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3$

Specific volume of the fluid :

$=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb$

Density of the fluid in British Gravitational System units = D $(slug/ft^3)$

1 kg = 0.06852 slug

1 m = 3.280 ft

$D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3$

Specific volume of the fluid :

$=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug$

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3 years ago