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2 years ago

Physical properties of earth metals-

Chemistry

1 answer:

Liula [17]2 years ago

4 0

Answer:

Properties of Alkaline Earth Metals

shiny.

silvery-white.

somewhat reactive metals at standard temperature and pressure.

readily lose their two outermost electrons to form cations with a 2+ charge.

low densities.

low melting points.

low boiling points.

Explanation:

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If we react 5.4g of sodium chloride with an unknown amount of fluorine gas, we produce 4.9g of sodium fluoride and 3.7g chlorine

Bumek [7]

Answer:

4.43 g

Explanation:

The reaction between sodium chloride and flourine gas is given as;

NaCl + F2 --> NaF + Cl2

From the stochiometry of the equation;

1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2

Mass of 1 mol of F2 = 38g

Mass of 1 mol of sodium flouride, NaF = 42g

This means 38g of flourine reacted with NaCl to form 42g of NaF

xg of F2 would form 4.9g of NaF

38 = 42

x = 4.9

x = 4.9 * 38 / 42

x = 4.43 g

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2 years ago

What are some of the main ways to add energy to a reaction? plzzz help

Tresset [83]

Answer:

Heat.

Light.

Sound.

Electricity

Explanation:

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2 years ago

What is the function of a restriction enzyme?

Advocard [28]

Main function of a restriction enzyme is to cleve (Cut) DNA bases.

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5 0

3 years ago

A star is estimated to have a mass of 2.0 x 10 ^36kg. Assuming it to be a sphere of average radius of 7.0 x 10 ^5 km. Calculate

Montano1993 [528]

Answer:

a) 1.392 x 10^6 g/cm^3

b) 8.69 x 10^7 lb/ft^3

Explanation:

mass of the star m = 2.0 x 10^36 kg

radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m

The density of substance ρ = mass/volume

The volume of the star = volume of a sphere = $\frac{4}{3}\pi r^{3}$

==> V = $\frac{4}{3}*3.142*(7.0*10^8)^{3}$ = 1.437 x 10^27 m^3

density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3

in g/cm^3 = (1.392 x 10^9)/1000 = 1.392 x 10^6 g/cm^3

in lb/ft^3 = (1.392 x 10^9)/16.018 = 8.69 x 10^7 lb/ft^3

6 0

2 years ago

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

6 0

2 years ago