CHEMISTRY COLLEGE

A solution of HNO3HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 2H++Na2CO3⟶2Na++H2O+CO2 A volume of 27.71±0.05 mL27.71±0.05 mL of HNO3HNO3 solution was required for complete reaction with 0.9585±0.0007 g0.9585±0.0007 g of Na2CO3Na2CO3 , (FM 105.988±0.001 g/mol105.988±0.001 g/mol ). Find the molarity of the HNO3HNO3 solution and its absolute uncertainty.

Answers

Answer 1

Answer:

Answer:

(0,653±0,002) M of HNO₃

Explanation:

The reaction of standarization of HNO₃ with Na₂CO₃ is:

2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻

To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:

0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =

9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃

Molarity is moles divide liters, thus, molarity of HNO₃ is:

1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃

The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:

ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =

0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M

Thus, molarity of HNO₃ solution and its absolute uncertainty is:

(0,653±0,002) M of HNO₃

I hope it helps!

Related Questions

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. srh2(s) + 2 h2o(l) sr(oh)2(s) + 2 h2(g) you wish to calculate the mass of hydrogen gas that can be prepared from 5.00 g of srh2 and 5.47 g of h2o. (a) how many moles of h2 can be produced from the given mass of srh2?
All the elements beyond uranium, the transuranium elements, have been prepared by bombardment and are not naturally occurring elements. The first transuranium element neptunium, NpNp, was prepared by bombarding U−238U−238 with neutrons to form a neptunium atom and a beta particle. Part A Complete the following equation: 10n+23892U→?+?01n+92238U→?+? Express your answer as a nuclear equation.
Write the full ground state electron configuration of f+.
If a student weighs out 0.744 g Fe ( NO 3 ) 3 ⋅ 9 H 2 O , what is the final concentration of the ∼0.2 M Fe ( NO 3 ) 3 solution that the student makes?
Which of the following is an acceptable structure for 2,5,5-trimethylhept-3-yne (CH3CH2)CH(CH3)C≡CCH2CH(CH3)2 CH3CH2C(CH3)2C≡CC(CH3)3 (CH3CH2)2C(CH3)C≡CCH2CH3 (CH3CH2)C(CH3)2C≡CCH(CH3)2 (CH3CH2CH2)CH(CH3)C≡CC(CH3)3

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=4.90 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

Answers

Answer : The concentration of $COF_2$ remains at equilibrium will be, 0.37 M

Explanation : Given,

Equilibrium constant = 4.90

Initial concentration of $COF_2$ = 2.00 M

The balanced equilibrium reaction is,

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

Initial conc. 2 M 0 0

At eqm. (2-2x) M x M x M

The expression of equilibrium constant for the reaction will be:

$K_c=([CO_2][CF_4])/([COF_2]^2)$

Now put all the values in this expression, we get :

$4.90=((x)* (x))/((2-2x)^2)$

By solving the term 'x' by quadratic equation, we get two value of 'x'.

$x=1.291M\text{ and }0.815M$

Now put the values of 'x' in concentration of $COF_2$ remains at equilibrium.

Concentration of $COF_2$ remains at equilibrium = $(2-2x)M=[2-2(1.219)]M=-0.582M$

Concentration of $COF_2$ remains at equilibrium = $(2-2x)M=[2-2(0.815)]M=0.37M$

From this we conclude that, the amount of substance can not be negative at equilibrium. So, the value of 'x' which is equal to 1.291 M is not considered.

Therefore, the concentration of $COF_2$ remains at equilibrium will be, 0.37 M

____________ acids are proton donors. ____________ bases are proton acceptors. The ____________ the pKₐ the stronger the acid. ____________ acids are electron pair acceptors.

Answers

Answer:

1. Bronsted—Lowry acid

2. Bronsted—Lowry Base

3. Lower the pka

4. Lewis acids

Explanation:

What makes physical change a physical change and a chemical change a chemical change

Answers

A physical change is any change in a substances form that does not change its chemical makeup. Examples of physical changes are breaking a stick or melting ice. A chemical change occurs when atoms of a substance are rearranged, and the bonds between the atoms are broken or formed. HOPE THIS HELPS!!

Answer:

A physical change is reversible and the changes made are only physical (smell, physical state, volume, etc) and most importantly no new substance is formed. In a chemical change, a new substance is formed, the chemical properties are changed, and it's permanent.

Show the conversion factor from Patolbf/ft2is 0.02089.

Answers

Explanation:

1 Pascal = 1 N/m²

To convert Pa to lbf/ft²

So, the conversion of N to pound force (lbf) is shown below as:

1 N = 0.224809 pound force (lbf)

The conversion of m² to ft² is shown below:

1 m² = 10.7639 ft²

So,

[tex]1\ Pa=\frac {1\ N}{1\ m^2}=\frac {0.224809\ lbf}{10.7639\ ft^2}

1 Pa = 0.02089 lbf / ft²

Hence proved.

Does the molarity of the solution change with the change in temperature?Match the words in the left column to the appropriate blanks in the sentences on the right.