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2 years ago

0.40 L of HNO3 is titrated to equivalence using 0.18 L of 0.1 M NaOH. What is the concentration of the HNO3 ?

1 answer:

3 0

This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.

<h3>Acid-base titrations:</h3>

In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:

$M_AV_A=M_BV_B$

Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.

Thus, we can write the reaction between NaOH and HNO3 as:

$HNO_3+NaOH\rightarrow NaNO_3+H_2O$

In such a way, we can solve for the concentration of the acid as shown below:

$M_A=\frac{M_BV_B}{V_A} \\\\M_A=\frac{0.1M*0.18L}{0.40L} \\\\M_A=0.045M$

Learn more about titration: brainly.com/question/25485091

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lisabon 2012 [21]

Answer:option

Option 5..IVA

Explanation:

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3 years ago

What is the molarity of solution obtained when 5.71 g of sodium carbonate-10-water is dissolved in water and made up to 250.0 cm

disa [49]

We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to $\frac{5.71}{286}$ = 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is $\frac{0.0199 X 1000}{250}$ = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³

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3 years ago

How many moles are in 5.03 x 1023 molecules?

AlladinOne [14]

Answer:

Some formulas for calculating mole are

Mole = Mass/ Molar mass

Mole = no of particles / avogadros constant

NB : no of particles can be no of atoms , no of ions , or no of molecules 2. Avogadros number or constant = 6.02 times 10 ^23

so we will be using the second formula

Mole = no of particles / avogadros constant

Mole = 5.03 x 10 ^23/6.02 x10^23

Mole = 8.355x10^45

hope it helps :)

Explanation:

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3 years ago

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Crazy boy [7]

26 because you need to do 54-28=26
The atomic mass number is the number of protons AND neutrons together.

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3 years ago

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3 years ago