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andre [41]
3 years ago
14

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?
Chemistry
2 answers:
Ber [7]3 years ago
8 0

Explanation:

Reactants are the species that are present on left hand side of a chemical reaction. Whereas products are the species that are present on right hand side of a chemical reaction.

For example, H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O

Here, both sulfuric acid and sodium hydroxide are the reactants. On the other hand, both water and sodium sulfate are the products.  

sattari [20]3 years ago
5 0
H2SO4 + 2NaOH = Na2SO4 + 2H20
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Ag + HCl --&gt; ?<br> a) No reaction<br> b) AgCl + H2<br> c) AgCl + H2O<br> PLEASE HELP
Volgvan

Answer:

AgCl + H2 - Chemical Equation Balancer.

8 0
3 years ago
Read 2 more answers
Automobile airbags contain solid sodium azide, NaN 3 , that reacts to produce nitrogen gas when heated, thus inflating the bag.
Yuri [45]

<u>Answer:</u> The value of work for the system is -935.23 J

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of sodium azide = 16.5 g

Molar mass of sodium azide = 65 g/mol

Putting values in above equation, we get:

\text{Moles of sodium azide}=\frac{16.5g}{65g/mol}=0.254mol

The given chemical equation follows:

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

By Stoichiometry of the reaction:

2 moles of sodium azide produces 3 moles of nitrogen gas

So, 0.254 moles of sodium azide will produce = \frac{3}{2}\times 0.254=0.381mol of nitrogen gas

To calculate volume of the gas given, we use the equation given by ideal gas, which follows:

PV=nRT

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = ?

T = Temperature of the gas = 22^oC=[22+273]K=295K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of nitrogen gas = 0.381 moles

Putting values in above equation, we get:

1.00atm\times V=0.381mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\V=\frac{0.381\times 0.0821\times 295}{1.00}=9.23L

To calculate the work done for expansion, we use the equation:

W=-P\Delta V

We are given:

P = pressure of the system = 1atm=1.01325\times 10^5Pa     (Conversion factor:  1 atm = 101325 Pa)

\Delta V = change in volume = 9.23L=9.23\times 10^{-3}m^3     (Conversion factor:  1m^3=1000L )

Putting values in above equation, we get:

W=-1.01325\times 10^5Pa\times 9.23\times 10^{-3}m^3\\\\W=-935.23J

Hence, the value of work for the system is -935.23 J

7 0
4 years ago
The first step in the formation of all three products is the loss of the ots leaving group to give a carbocation intermediate. t
lina2011 [118]

The carbocation stabilized by resonance structure and thereby lowers the energy of the carbocation, hydrogen will add to the carbon in the double bond that produces delocalization of electrons.

<h3>What is carbocation?</h3>

A carbocation is a molecule in which a carbon atom has a positive charge and three bonds.

In general, electrons are stabilized by delocalization. The stabilization energy engendered by delocalization over more than two atoms is called the resonance stabilization energy or simply the resonance energy. The greater the extent of electron delocalization the greater the resonance stabilization.

Learn more about the carbocation here:

brainly.com/question/19168427

#SPJ1

4 0
2 years ago
HCIO4 + P2O10 = 4 H3PO4 + CL207 is this balanced or unbalanced
Inessa05 [86]
Right Side:
H- 1
Cl- 1
O- 4+10=14
P- 2

Left Side:
H- 12
P- 4
O- 4*4+7=23
Cl- 2

The equation is unbalanced.
5 0
3 years ago
g A 500. mL solution contains 0.665 M NaC2H3O2 and 0.475 M HC2H3O2. What mass of HCl in grams needs to be added for the solution
77julia77 [94]

Answer:

7.38g HCl

Explanation:

Using H-H equation for acetic buffer:

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

<em>Where pKa is -log Ka = 4.74 and [] could be taken as moles of each compound.</em>

The initial moles of each specie is:

[NaC2H3O2]:

0.500L * (0.665mol/L) = 0.3325moles

[HC2H3O2]:

0.500L * (0.475mol/L) = 0.2375 moles

That means total moles are:

[NaC2H3O2] + [HC2H3O2] = 0.57 moles <em>(1)</em>

And solving H-H equation for a pH of 4.21:

4.21 = 4.74 + log [NaC2H3O2] / [HC2H3O2]

0.29512 = [NaC2H3O2] / [HC2H3O2] <em>(2)</em>

Replacing (1) in (2):

0.29512 = 0.57mol - [HC2H3O2] / [HC2H3O2]

0.29512 [HC2H3O2] = 0.57mol - [HC2H3O2]

1.29512 [HC2H3O2] = 0.57mol

[HC2H3O2] = 0.44 moles

The HCl reacts with NaC2H3O2 producing HC2H3O2, that means you need to add:

0.44 moles - 0.2375 moles =

0.2025 moles of HCl

Using molar mass of HCl (36.45g/mol), to convert these moles to grams:

0.2025 moles * (36.45g/mol) =

<h3>7.38g HCl</h3>

8 0
3 years ago
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