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3 years ago

A reaction in which two ions switch places is which type of reaction?

Chemistry

1 answer:

AURORKA [14]3 years ago

6 0

A reaction in which two ions switch places is called double-displacement reaction.

Explanation:

Double displacement reaction usually takes place between two positive ions or two negative ions. When anions or cations of totally different compounds change places, two totally different compounds are formed.

<u>Syntax for double displacement reaction: </u>

AB + CD --> AD + BC

It can also be defined in simple terms as the reaction in which exchange of ions takes place is called double displacement reaction.

The chemical reaction is as follows:

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

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A fuel gas containing 86% methane, 8% ethane, and 6% propane by volume flows to a furnace at a rate of 1450 m3/h at 15°C and 150

Svetach [21]

Answer:

Basis: Hour

From the question, we will have the following reactions;

CH4 + 2O2 -------- CO2 + 2H20 (Methane with O2)

C2H6 + 3.5O2 -------- 2CO2 + 2H2O (Butane with O2)

C3H8 + 5O2 ---------- 2CO2 + 4H2O (Propane with O2)

But we are also given this,

R=8.314J/mol.K, V=1450m3/h, P=150kPa gauge, Pt=150+101kPa=251kPa, T=15C= 288K

Assuming they are all ideal gases, we can find the no of moles of the gases.

n=PV/RT,

n = 251x103 x 1450 /8.314 x 288

n = 151, 999mols = 152kmols

However from the input and complete reactions stoichiometries above, we will have,

1. Methane 86% = 0.86 x 152kmols = 130kmols, required O2 = 2 x 130.7 = 261.44kmols

2. Ethane 8% = 0.08 x 152kmols = 12.2kmols, required O2 = 3.5 x 12.2 = 42.56kmols

3. Propane 6% = 0.06 x 152 kmols = 9.2kmols, required O2 = 5 x 9.1 = 45.5kmols

O2 = 349.5kmols, with 8% excess, Total O2 = 349.5+ (0.08x349.5) = 377.46kmols

But Air:O2 = 21%: 100%

inflow Air = 377.46x 100/21= 1797.5kmols, at standard pressure and temperature.

From PV =nRT

V (M3/H) = nRT/P

===== 1797.5mol x 8.314Nm/mol.K x 273K/101325Nm-2 x 1000

Hence, the required flow rate of air in SCMH = 40,265m³/h

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Read 2 more answers

Both diamond and graphite (i.e. pencil lead) consist of carbon atoms. They are only different in their crystalline structures. O

prisoha [69]

Answer:

Moles of carbon atoms = 3.33 × $10^{-6}$ mol

No. of atoms of C in Diamond = 2.007 × $10^{28}$ atom

Atoms of graphite = 6.27 × $10^{23}$ Atoms

Explanation:

given data

Cost of 0.2g of diamond = $5000

Cost of 25 g of graphite = $ 2

solution

we know cost of 0.2g of diamond is $ 5000 so that for 1$

if buy 1$ = $\frac{0.20}{5000}$

1$ = 4.0 × $10^{-5}$ g Carbon

and Moles of carbon atoms is express as

Moles of carbon atoms = Given mass of Carbon ÷ atomic mass of C .........1

Moles of carbon atoms = 4.0 × $10^{-5}$ g/ 2.0g

Moles of carbon atoms = 3.33 × $10^{-6}$ mol

and

No. of atoms of C in Diamond = No. of moles × Avogadro NO ..............2

No. of atoms of C in Diamond = 3.33 × $10^{-6}$ mol × 6.022 × $10^{28}$

No. of atoms of C in Diamond = 2.007 × $10^{28}$ atom

Graphite

and wew have given Cost of 25 g of graphite is $2 so for but 1$ we get

for buy $1 = 25÷2 = 12.5 g Of graphite

Moles of graphite = 12.5÷12 = 1.04 mol

Atoms of graphite = 1.04 × 6.022 × 1023

Atoms of graphite = 6.27 × $10^{23}$ Atoms

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3 years ago