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lisabon 2012 [21]
2 years ago
10

What is the formula equation for the reaction between sulfuric acid and dissolved sodium hydroxide if all products and reactants

are in the aqueous or liquid phase?
mc007-1.jpg
mc007-2.jpg
Chemistry
1 answer:
Mnenie [13.5K]2 years ago
3 0

Answer : The formula equation will be:

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

When sulfuric acid react with sodium hydroxide then it react to give sodium sulfate and water as a product.

The balanced chemical reaction will be:

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

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mihalych1998 [28]
This isn’t anything related to chemistry dude
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3 years ago
Help which one will it be
salantis [7]

Answer:

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Explanation:

6 0
2 years ago
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1. Imagine you have a sealed 20.0 L balloon filled with helium gas at 750 mmHg in the
SVETLANKA909090 [29]

Answer:

17.6510 L

Explanation:

First we should get the number of moles of helium here by Boyle's law

PV=nRT

P=750/760= 0.9868 atm

T=25+273=298 kelvin

R= 0.08206

V= 20L

so

n=PV/RT

n=0.9868×20/0.08206×298

n=0.80707 mol

Then use the same law

V=0.80707×0.08206×263/0.9869=

17.6510L

SO THE VOLUME WILL BE 17.6510 L

6 0
2 years ago
How many atoms of mercury are present in 3.2 cubic centimeters of liquid mercury? the density of mercury is 13.55 g/cc. answer i
jenyasd209 [6]

Answer:

             1.3 × 10²³ Atoms of Mercury  

Solution:

Step 1: Calculate Mass of Mercury using following formula,

                               Density  =  Mass ÷ Volume

Solving for Mass,

                               Mass  =  Density × Volume

Putting values,

                               Mass  =  13.55 g.cm⁻³ × 3.2 cm³                ∴ 1 cm³ = 1 cc

                               Mass  =  43.36 g

Step 2: Calculating number of Moles using following formula;

                               Moles  =  Mass ÷ M.mass

Putting values,

                               Moles  =  43.36 g ÷ 200.59 g.mol⁻¹

                               Moles  =  0.216 mol

Step 3: Calculating Number of Atoms using following formula;

                               Number of atoms  =  Moles × 6.022 ×10²³

Putting value of moles,

                               Number of Atoms  =  0.216 mol × 6.022 × 10²³

                              Number of Atoms  =  1.3 × 10²³ Atoms of Hg

7 0
3 years ago
Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2
Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has SP^{2}  O has SP^{2} .

Explanation : The orbitals of oxygen and carbon which are involved in SP^{2} hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and  O also has π  orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b:  O SP^{3}  H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen SP^{3} hydrogen atoms in it. It has SP^{3} hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c:   C is SP^{3}  O is also SP^{3}

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of SP^{3} hybridized bonds.

Answer 5) Bond d:   C atom has SP^{3}  C atom has SP^{3}

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be SP^{3}. Therefore, the orbitals that undergo SP^{3} hybridization are SP^{3}.

Answer 6) Bond e : C1 containing O SP^{2}    

C2 is SP^{3}

Explanation : The carbon atom which contains oxygen along with a double bond has SP^{2} hybridized orbitals involved in the bonding process; whereas the carbon at C2 has SP^{3} hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0
3 years ago
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