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Vikentia [17]
3 years ago
7

Which of the following represents a neutralization reaction which hydrofluoric acid and sodium hydroxide(a base) react to produc

e sodium choride(a salt) and water? 
Chemistry
1 answer:
Olenka [21]3 years ago
3 0

Explanation:

A neutralization reaction is a reaction that involves an acid and a base.

When an acid reacts with a base to produce salt and water only, the reaction is said to be a neutralization reaction.

  Acid    +  Base    →  Salt  +  Water

Such reactions are classified as a double displacement reaction in which a non-ionizing product is formed

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Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu
IrinaK [193]

Answer : The atomic radius for Ti is, 1.45\times 10^{-8}cm

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number (N_{A})=6.022\times 10^{23} mol^{-1}

First we have to calculate the volume of HCP crystal structure.

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times V} .............(1)

where,

\rho = density  = 4.51g/cm^3

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass  = 47.87 g/mole

(N_{A}) = Avogadro's number  

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}

V=1.06\times 10^{-22}cm^3

Now we have to calculate the atomic radius for Ti.

Formula used :

V=6R^2c\sqrt{3}

Given:

c/a ratio = 1.669 that means,  c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

V=6R^2\times (1.669a)\sqrt{3}

V=6R^2\times (1.669\times 2R)\sqrt{3}

V=(1.669)\times (12\sqrt{3})R^3

Now put all the given values in this formula, we get:

1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3

R=1.45\times 10^{-8}cm

Therefore, the atomic radius for Ti is, 1.45\times 10^{-8}cm

3 0
3 years ago
What is the pressure of a gas that began at 38 torr, and 500L and is changed to occupy a volume of 677 L?
qwelly [4]

Answer:

P₂ = 28.5 torr

Explanation:

Given data:

Initial pressure = 38 torr

Initial volume = 500 L

Final volume = 677 L

Final pressure = ?

Solution:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = Initial volume

P₂ = Final pressure

V₂ = Final volume

Now we will put the vales in formula.

P₁V₁ = P₂V₂

P₂ = P₁V₁ /V₂

P₂ = 38 torr × 500 L / 667 L

P₂ = 19000 torr. L / 667 L

P₂ = 28.5 torr

6 0
3 years ago
Express the rate of reaction in terms of the change in concentration of each of the reactants and products. 2A(g) → B(g) + C(g)
ankoles [38]
[B][C] / [A]^2

Products raised to the coefficients over reactants raised to the coefficients
8 0
3 years ago
An ordered list of chemical substances is shown. Chemical Substances 1 CuO 2 O2 3 CO2 4 NO2 5 Fe 6 H2O Which substances in the l
a_sh-v [17]

Answer:

reactants: 2 O2

products: 3 CO2, 4 NO2, 6 H2O

Explanation:

In a combustion, a combustible material, which generally is composed of C, H, O, N, and S, is combusted, that is, react with oxygen after a spark was produced; obtaining fire, heat and subproducts, including ashes and gases.

Oxygen is always one of the reactants of a combustion.

If Nitrogen was present in the combustible, NO2 (or other nitrogen oxides) will be produced.

If Carbon was present in the combustible, CO2 will be produced (also CO can be produced).

If Hydrogen was present in the combustible, H2O will be produced.

7 0
3 years ago
If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your
Drupady [299]

N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

<em>N₂+3H₂⇒2NH₃</em>

mol N₂(MW=28 g/mol) :

\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571

mol H₂(MW= 2 g/mol) :

\tt mol=\dfrac{100}{2}=50

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

\tt \dfrac{3.571}{1}\div \dfrac{50}{3}=3.571\div 16.6

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant

5 0
2 years ago
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