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3 years ago

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Explain what happens when water reacts with sodium metal. Support your answer with the relevant

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Sodium metal reacts rapidly with water to form a solution of sodium hydroxide (NaOH) and hydrogen gas (H2). This reaction is exothermic.

Equation:

2Na + 2H²0 --------}- 2NaOH + H²

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Consider the following reaction:

NeX [460]

For the given reaction, according to the Law of Conservation of Energy, the energy required to decompose Hcl and produce $H_{2}+c l_{2}$ are equal.

Answer: Option C

<u>Explanation:</u>

According to law of conservation's of energy, energy can only be transferred from reactants to product side. So in this process, it is stated that 185 kJ of energy will be needed to decompose it. So that 185 kJ of energy will be getting transferred to produce the creation of hydrogen and chloride in the product side.

So if we see from the reactants side, the energy of 185 kJ is required for decomposition of hydrogen chloride. Similarly, if we see from the product side, the 185 kJ utilized for decomposition is transferred as energy required to create hydrogen and chlorine atoms. This statement will be in accordance with the law of conservation's of energy.

3 0

3 years ago

Read 2 more answers

2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0

3 years ago

A recently depolarized area of a cell membrane cannot generate an action potential because of the

marysya [2.9K]

The refractory period makes the cell rest. In this period, the cell becomes able to depolarize again by collecting ions.

4 0

3 years ago

Read 2 more answers

Fundamental differences in the cell, the basic unit of life, allow for broadest classification of all living organisms into thre

Nana76 [90]

Answer:

C

Explanation:

7 0

3 years ago

An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu

chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\pi r^3$ ..[1]

Diameter of the nucleus ,d' = $9.00\times 10^{-5}\AA$

Radius of the nucleus ,r' = 0.5 d'= $0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA$

Volume of nucleus = V'

$V=\frac{4}{3}\pi r'^3$ ..[2]

Dividing [2] by [1]

$\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}$

$=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}$

$\frac{V'}{V}=4.6656\times 10^{-14}$

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

Diameter of the proton ,d = $1.72\times 10^{-15} m = 1.72\times 10^{-13} cm$

1 m = 100 cm

Radius of the proton,r = 0.5 d= $0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm$

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3$

Mass of proton, m = 1.0073 amu = $1.0073\times 1.66054\times 10^{-24} g$

$1 amu = 1.66054\times 10^{-24} g$

Density of the proton : d

$d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3$

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

5 0

3 years ago