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3 years ago

What is the formula for CO3 2- and Al3+ ?

Chemistry

2 answers:

Goryan [66]3 years ago

7 0

Answer is: formula is Al₂(CO₃)₃.

Aluminium carbonate (Al₂(CO₃)₃) has neutral charge. Because aluminium cation has positive charge 3+ and carbonate anion has negative charge 2-, for right chemical formula, we need two aluminium cations and three carbonate anion:

charge of the molecule = 2 · (3+) + 3 · (-2).

charge of the molecule = 0.

blsea [12.9K]3 years ago

5 0

The formula for the given above is obtained by cross multiplying the charges of the given set of atoms. The element with a positive value for the charge should be written first. This may be shown through the formula,
Al2(CO3)2.

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What elements do not fit the pattern of electron affinity

Pavel [41]

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3 years ago

A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mix

Alisiya [41]

Answer:

$M_f=38.8\%$

Explanation:

From the question we are told that:

Pressure $P=747mmHg$

Temperature $T=298K$

Volume $V=11.1$

Heat Produced $Q=780kJ$

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

$n= (747/760) *11.1/ (0.0821*298)$

$n=0.446mol$

Therefore

$x+y=0.446$

$x=0.446-y .....1$

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

$x(889) + y(2220) = 760 ...... 2$

Comparing Equation 1 and 2 and solving simultaneously

$x=0.446-y .....1$

$x(889) + y(2220) = 760 ...... 2$

$x=0.173$

$y=0.273$

Therefore

Mole fraction 0f Methane is mathematically given as

$M_f=\frac{x}{n}*100\%$

$M_f=\frac{1.173}{0.446}*100\%$

$M_f=38.8\%$

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3 years ago

Is diphenylamine a solid, liquid, or a gas at room temperature

goblinko [34]

Answer:

Solid

Explanation:

Diphenylamine has a melting point of 127.4 F or 53 C so at room temperature ~70 F or 21 C its a solid.

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3 years ago

What would limit the population growth of rabbits in a forest ecosystem?

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Other predators? or other factors that affect the food chain ie extinction of whatever source the rabbits feed off of

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3 years ago

How many grams of aluminum chloride are needed to react completely with 1.084g lithium sulfide?

Oksi-84 [34.3K]

Molar mass :

Li₂S = 45.947 g/mol

AlCl₃ = 133.34 g/mol

3 Li₂S + 2 AlCl₃ = 6 LiCl + Al₂S₃

3 * 45.947 g Li₂S ----------> 2 * 133.34 g AlCl₃
1.084 g Li₂S ----------------> ?

Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947

Mass Li₂S = 289.08112 / 137.841

Mass Li₂S = 2.0972 g

hope this helps!

5 0

2 years ago