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4 years ago

The process n01 → p11 + β−10 n 0 1 → p 1 1 + β - 1 0 represents a(n) _____.

Chemistry

1 answer:

polet [3.4K]4 years ago

5 0

Answer:

beta decay

Explanation:

The process is represented by the nuclear equation:

$^1_0n$ → $^1_1p+^0_{-1}\beta$

Where:

n represents a neutron,
p represents a proton, and
β represents an electron.

The superscripts to the leff of each symbol is the mass number (number of protons and neutrons), and the subscript to the left means the atomic number (number of protons).

Then, in this process a neutron is being transformed into a proton by the emssion of an electron (from inside the nucleus of the atom).

This electron is named beta (β) particle, and the process is called beta decay, because the neutron is changing into other subatomic particles.

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What is a molarity a measurement

USPshnik [31]

Answer:

Molarity indicates the number of moles of solute per liter of solution and is one of the most common units used to measure the concentration of a solution.

Explanation:

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4 years ago

Differentiate between group and period

Katarina [22]

Answer:

A group is a vertical column of elements on the periodic table whereas a period is a horizontal row of elements on the periodic table.

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3 years ago

Read 2 more answers

The equilibrium constant is given for one of the reactions below.

Cerrena [4.2K]

Answer:

12.7551

Explanation:

The given chemical equation follows:

$2HD(g)\rightarrow H_2(g)+D_2(g)$

The equilibrium constant for the above equation is 0.28.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$2H_2(g)+2D_2(g)\rightarrow 4HD(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(\frac{1}{0.28})^2$

Hence, the value of equilibrium constant for reverse reaction is 12.7551.

3 0

4 years ago

If the ka of a monoprotic weak acid is 5.4 × 10-6, what is the ph of a 0.14 m solution of this acid?

Aleks [24]

pH = 2.1

Let $HA (aq)$ resembles the acid; as a weak acid (a small value of $K_{a}$ ) $HA$ would partially dissociate to produce protons $H^{+}$ and $A^{-}$ , its conjugate base. Let the final proton concentration (i.e., $[H^{+}]$ ) be $x$ . (Apparently $x \ge 0$ ) Construct the following RICE table:

$\left\begin{array}{cccccc}\text{R}&HA(aq)&\rightleftharpoons&H^{+}(aq) &+ &A^{-}(aq)\\\text{I} & 0.14 \; \text{M} & &\\\text{C}&-x \; \text{M}& &+x \; \text{M} & & +x \; \text{M}\\E & (0.14 - x)\; \text{M} & & x \; \text{M} & &\+x \; \text{M}\end{array}\right$

By definition, (all concentrations are under equilibrium condition)

$\left\begin{array}{ccc}K_{a}&=&[H^{+}] \cdot [A^{-}] / [HA]\\&=&x^{2} /(0.14 - x)\end{array}\right$

It is given that

$K_a = 5.4 \cdot 10^{-6}$

Equating and simplifying the two expressions gives a quadratic equation; solve the equation for $x$ gives:

$x^2 = 5.4 \cdot 10^{-6} \cdot (14 - x) \\x^2 + 5.4 \cdot 10^{-6} \cdot 14 \cdot x - 5.4 \cdot 10^{-6} \cdot 14 = 0 \\x = 0.0087 \; \text{M} \; (x \ge 0)$

The pH of a solutions equals the opposite of the logarithm of its proton concentration to base 10; thus for this particular solution

$\text{pH} = -\text{ln(}[H^{+}]\text{)} / \text{ln(}10\text{)} = 2.1$

7 0

4 years ago

Why does silicon in silicon dioxide bond with 4 oxygen atoms and not 2?

Elis [28]

Answer:

Silicon does not form double bonds with oxygen, whereas carbon is capable of forming double bonds with oxygen. While the carbon dioxide molecular structure is linear, the silicon dioxide has an extended, different covalent structure.

Explanation:

If the sizes of the atoms of Silicon (Si) and Carbon (C) are compared to each other, the Si atoms are larger than carbon - which implies that the Si-O bonds will be longer than the C-O bonds. As a result, the p orbitals present on the Si and O atoms aren't very near to each other, in order to get together for the required overlap sideways which could have formed a stable pi bond. Hence, Silicon forms only single covalent bonds with Oxygen in silicon dioxide, in the form of a diamond structure with each Si atom being connected to its four neighbouring atoms through an O atom.

On the other hand, in the case of carbon dioxide, C is perfectly capable of forming double bonds with O. The different p orbitals are brought close together, resulting in a sideways overlap that leads to two pi bonds, twisted at a right angle to each other. As a result, the Carbon in carbon dioxide bonds with 2 oxygen atoms but not 4.

5 0

4 years ago