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lesya692 [45]
3 years ago
9

What do the subscripts mean in a molecular formula ?​

Chemistry
2 answers:
Zigmanuir [339]3 years ago
8 0

Answer:

Explanation:

Subscripts. We use subscripts in chemical formulae to indicate the number of atoms of an element present in a molecule or formula unit.

horsena [70]3 years ago
3 0

Answer:

A molecular formula tells us what atoms and how many of each type of atom are present in a molecule.

Explanation:

If only one atom of a specific type is present, no subscript is used.

For atoms that have two or more present, a subscript is written after the symbol for that atom.

Molecular formulas do not indicate how the atoms are arranged in the molecule.

- Hope this helps! -blazetoxic123

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4 years ago
g What is the mass ratio for CaCl2 if a sample analyzed to be 36.1 grams calcium and 63.9 grams chlorine
Kitty [74]

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0.361Ca\\\\ 0.639Cl

Explanation:

Hello!

In this case, since the mass ratio here can be computed as:

\frac{mass\ Ca}{mass\ CaCl_2} \\\\\frac{mass\ Cl}{mass\ CaCl_2}

For each element forming the compound we obtain:

\frac{36.1g}{63.9g+36.1g} = 0.361Ca\\\\\frac{63.9g}{63.9g+36.1g} = 0.639Cl

Best regards!

7 0
3 years ago
At 920 K, Kp = 0.40 for the following reaction. 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g) Calculate the equilibrium p
Alex777 [14]

<u>Answer:</u> The equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

<u>Explanation:</u>

We are given:

Initial partial pressure of sulfur dioxide = 0.52 atm

Initial partial pressure of oxygen = 0.52 atm

Initial partial pressure of sulfur trioxide = 0 atm

For the given chemical reaction:

                        2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

<u>Initial:</u>                 0.52      0.52

<u>At eqllm:</u>         0.52-2x    0.52-x        2x

The expression of K_p for above equation follows:

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times p_{O_2}}

We are given:

K_p=0.40\\\\p_{SO_3}=2x\\\\p_{SO_2}=0.52-2x\\\\p_{O_2}=0.52-x

Putting values in above equation, we get:

0.40=\frac{(2x)^2}{(0.52-2x)^2\times (0.52-x)}\\\\x=-1.13,-0.402,0.077

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide = (0.52-2x)=(0.52-(2\times 0.077))=0.366atm

Equilibrium partial pressure of oxygen = (0.52-x)=(0.52-0.077)=0.443atm

Equilibrium partial pressure of sulfur trioxide = 2x=(2\times 0.077)=0.154atm

Hence, the equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

4 0
3 years ago
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