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3 years ago

1A, 3B, and 7A are examples of __________ on the periodic table.

Physics

2 answers:

SVEN [57.7K]3 years ago

5 0

Answer:

group number

Explanation:

nalin [4]3 years ago

3 0

Answer:

1 A, 3 B, and 7 A are examples of group number on the periodic table.

Explanation:

The table that consists of elements arranged systematically on the basis of atomic numbers is called periodic table. Groups and periods are two important part of a periodic table. A periodic table contains seven horizontal rows, called as periods and 18 vertical columns, called group.

The groups name is given in the form of number and alphabets. It represents the columns present in the periodic table.

Hence, 1 A, 3 B, and 7 A are examples of group number on the periodic table.

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In a Young’s interference experiment, the two slits are separated by 0.150 mm and the incident light includes two wavelengths: l

dezoksy [38]

Answer:

2.52 × 10⁻² cm

Explanation:

The distance of bright fringe from the center of the screen is given by the formula

$y = \frac{m\lambda D}{d}$

Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the

slits is d.

Separation between the slits, d = 0.15 mm

$= 0.15 * 10^{-3} m$

Distance of the screen from the slits = 1.40 m

We have a wavelength, λ1 = 540 nm

= $540 * 10^{-9} m$

By substituting all these values in the above equation we get

y1 = mλD/d

$y1 = m(540 \times 10^{-9} m)(1.40 m)/(0.15 \times 10^{-3} m)$

$y1 = m(5.04 * 10^{-3} m)$

We have a wavelength, λ2 = 450 nm

= $450 * 10^-9 m$

By substituting all these values in the above equation we get

$y_2 = \frac{m\lambda D}{d}$

$y_2 = m(450 * 10^{-9} m)(1.40 m)/(0.15 * 10^{-3} m)$

$y_2 = m'(4.20 * 10^{-3} m)$

According to the problem, these two distance are coincides with each other.

So,

$y_1 = y_2$

$m(5.04 * 10^{-3} m) = m'(4.20 * 10^{-3} m)$

by testing values, the above equation is satisfied only when, m = 5 and m' = 6

Then from the above we have

y1 = y2 = 0.0252 m

= 2.52 × 10⁻² cm

8 0

3 years ago

What is the ph of a buffer solution that is 0.172 m in hclo and 0.131 m in naclo? hint: the ka of hclo is 3.8 x 10-8.

nata0808 [166]

The pH of the buffer solution is 7.30 for 0.172 m in Hypochlorous acid and 0.131 m in Sodium hypochlorite.

<h3>What is a buffer solution?</h3>

A weak acid and its conjugate base, or a weak base and its conjugate acid, are mixed together to form a solution called a buffer solution, which is based on water as the solvent. They do not change in pH when diluted or when modest amounts of acid or alkali are added to them. A relatively little amount of a strong acid or strong base has little effect on the pH of buffer solutions. As a result, they are employed to maintain a steady pH.

According to the question:

Ka = 3.8×10⁻⁸

pKa = - log (Ka)

= - log(3.8×10⁻⁸)

= 7.42

pH = pKa + log {[conjugate base]/[acid]}

= 7.42+ log {0.131/0.172}

= 7.302

To know more about buffer solutions, visit:

brainly.com/question/24262133

#SPJ4

5 0