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3 years ago

In which of the ions do all x-o bonds (x indicates the central atom) have the same length?

Chemistry

1 answer:

Radda [10]3 years ago

7 0

<span>Answer: All of the ions do all x-o bonds (x indicates the central atom) have the same length.</span>

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Some versions of the periodic table show hydrogen at the top of Group 1A(1) and at the top of Group 7A(17). What properties of h

lawyer [7]

Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.

Explanation:

The reason why hydrogen is similar to group 1 metals:

#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion

#It acts as a good reducing agent similar to group1 metals

#It can also halides

Similarity to halogens:

#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.

#It exists as diatomin molecule,H2

#Have the same electronegativity nature

#its reaction with other metal

8 0

3 years ago

True or false: The Earth's rotation causes the winds to curve.

Rashid [163]

Answer:

true

Explanation:

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2 years ago

If the gas in the above system has a volume of 256 mL at a pressure

evablogger [386]

Answer:

the volume of the same gas at pressure of 1.00atm =737.3ml

6 0

2 years ago

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10

kkurt [141]

Answer:

The diameter of the hydrogen $\mathbf{d =1.0605 \times 10^{-10}\ m}$

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

$L = \dfrac{nh}{2 \pi}$

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

$mvr= \dfrac{nh}{2 \pi}$

$v= \dfrac{nh}{2 \pi mr}$ ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

$\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}$ ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

$\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2$

$ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}$

$r =\dfrac{n^2h^2}{4 \pi^2 mke^2}$

For ground-state n = 1

$h = (6.625 \times 10^{-34} \ J.s)^2$

$m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)$

$Ke = (1.6 \times 10^{-19} \ C)^2$

$r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}$

$r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}$

$\mathbf{r = 5.3025 \times 10^{-11} \ m}$

Therefore, the diameter of hydrogen d = 2r

$\mathbf{d = ( 2 \times 5.3025 \times 10^{-11} \ m})}$

$\mathbf{d =1.0605 \times 10^{-10}\ m}}$

4 0

3 years ago

Which 2 elements when combined NOT alloys?

Volgvan

Answer:

Oxygen and sulfur

Explanation:

Alloys consist of metals fused together sometimes with additional components, such as carbon, to prevent metals from corrosion.

Oxygen and sulfur are two compounds consisting of non-metal atoms, O and S. Combining oxygen with sulfur would actually produce sulfur dioxide, a gas, in contrast to a metallic substance that is fused with some other metal or carbon.

That's the reason why combination of sulfur and oxygen wouldn't produce an alloy: it would produce a gas.

6 0

3 years ago