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3 years ago

12

What is the formula for the conjugate base formed by the following acid HCN

1 answer:

meriva3 years ago

5 0

Answer:

HCN - Hydrogen cyanide The conjugate acid of CN- is HCN. HCN stands for hydrogen cyanide. The conjugate acids are a combination of a strong acid and a low base.

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What is the mass of H2O produced when 10.0 grams of H2 reacts completely with 80.0 grams of O2?

fenix001 [56]

Molar mass of O2= 16.0 g/mol
molar mass of H2O= 18.0 g/mol

10.0 g of O2= 5 mol
80.0 g of H20= 5 mol

you need twice as the moles of H2 as O2 so there is insufficient H2 and it is the limiting reactant and all of it will react therefore 5.00 moles of H2 with 2.5 mole of O2 to produce 5.00 moles H2O

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so the answer is 90.00

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4 years ago

How many grams of lithium nitrate, LiNO3 (68.9 g/mol), are required to prepare 342.6 mL of a 0.783 M LiNO3 solution ? A) 0.00389

Anika [276]

Answer:

b) 18.5 g

Explanation:

The first step is the <u>calculation of number of moles</u> of $LiNO_3$ with the molarity equation, so:

$M=\frac{#~mol}{L}$

With the <u>volume in "L" units</u> (342.6 mL= 0.342 L ) we can <u>find the moles</u>, so:

$0.783~M\frac{#~mol}{0.342~L}$

$#~mol=~0.783*0.342=0.268~mol~LiNO_3$

Now, we have to calculate the <u>molar mass</u> to convert the moles to grams. For this we have to know the <u>atomic mass</u> of each atom:

Li ( 6.94 g/mol)

N (14 g/mol

O (16 g/mol)

With these values and the number of atoms in the molecule we can do the math:

(6.94 x 1) + ( 14 x 1) + (16 x 3 ) = 68.94 g/mol

The final step is the <u>calculation of the grams</u>, so:

$0.268~mol~LiNO_3~\frac{69.94~g~LiNO_3}{1~mol~LiNO_3}=~18.5~g~LiNO_3$

<u>We will need 18.5 g of LiNO3 to do a 0.783 M solution with a volume 342.6 mL.</u>

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