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Zolol [24]
3 years ago
8

The flask contains 10.0mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.320 M NaOH. What is the init

ial concentration of HCl
Chemistry
1 answer:
yan [13]3 years ago
5 0

This question is incomplete, the complete question is;

The flask contains 10.0mL of HCl and a few drops of phenolphthalein indicator.

The buret contains 0.320 M NaOH,

It requires 11.2 mL of the NaOH solution to reach the end point of the titration

What is the initial concentration of HCI?

Answer:  

initial concentration of HCl is 0.3584 M

Explanation:

Given that;

the reaction between HCl and NaOH is 1:1 molar

HCl(aq) + NaOH(aq) -------> NaCl(aq) + H2O(l)

we know that molarity is the number of moles of solute per liter of solution

so moles of NaOH consumed

= ( 0.320 mol / 1000ml) × 11.2ml

= 0.003584 mol

now

0.003584 moles of NaOH reacts with 0.003584 moles of HCl

moles of HCl present in the 10.0ml solution = 0.003584 mol

therefore the initial concentration of HCl

= ( 0.003584 mol / 10.0ml )× 1000ml

= 0.3584 M

the initial concentration of HCl is 0.3584 M

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The volume of the earth is approximately 1.08326x10 to the 12th power km to the 3rd power. How many liters is this
mafiozo [28]

Here, we are required to determine the volume of the earth which is 1.08326 × 10¹² km³ in liters.

<em>The volume of the earth is approximately</em>,

, 1.08326 × 10²⁴ liters

By conversion factors;

  • <em>1dm³ = 1liter</em>
  • However; <em>1km = 10000dm = 10⁴ </em><em>dm</em>
  • Therefore, 1km³ = (10⁴)³ dm³.

Consequently, 1km³ = 10¹²dm³ = 10¹²liters.

The conversion factor from 1km³ to liters is therefore, c.f = 10¹²liters/km³

Therefore, the volume of the earth which is approximately, 1.08326 × 10¹² km³ can be expressed in liters as;

<em>1.08326 × 10¹² km³ × 10¹²liters/km³ </em>

The volume of the earth is approximately,

1.08326 × 10²⁴ liters.

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7 0
2 years ago
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

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From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

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Define the term hydrate
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