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3 years ago

What orbital type is being filled throughout the transition metal B Groups?

Chemistry

1 answer:

Nataliya [291]3 years ago

8 0

Answer:

The correct answer is: d-orbital

Explanation:

The transition metals are the chemical elements that belong to the B-group of the periodic table, from group 1B to 8B. The B-group is located between the groups IIA and IIIA in the periodic table.

The general electronic configuration of the chemical elements belonging to this group is (n-1) d¹⁻¹⁰ ns⁰⁻², because of the <u>partially filled d-subshell</u> in the ground state or excited state. <u>Therefore, the transition metals are refereed as the </u><u>d-block elements</u><u>.</u>

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A rectangular fence is 3 times as long as it is wide. The perimeter is 60ft. Find the dimensions of the fence

Mashutka [201]

Answer:

The length of the fence is 22.5ft and the width of the fence is 7.5ft.

Explanation:

To solve this problem we will turn this question into an equation.

The equation for the perimeter of a rectangle is 2l + 2w = perimeter.

In the question we were given the perimeter so we can go ahead and plug this in.

2l + 2w = 60

In the question we are also told that the fence is 3 times as long as it is wide. This means that 1l = 3w. We can now subsitute l for 3w so that we only have one variable.

2(3w) + 2w = 60

Now we can simply use algebra to solve for w.

6w + 2w = 60

8w = 60

w = 7.5

Now that we know the value of w we can find the value of l since we know 1l = 3w.

l = 3w

l = 3(7.5)

l = 22.5

The length of the fence is 22.5ft and the width of the fence is 7.5ft.

7 0

3 years ago

What did Ernest Rutherford s gold foil experiment demonstrate about atoms?

mote1985 [20]

Rutherford performed gold foil experiment to understand that how negative and positive particles could Co exist in an atom. He bombarded alpha particles on a 0.00004 cm thick gold foil.

He proposed a planetary model of the atom and concluded following results and demonstrated that,
1. An atom produces a line spectrum.
2. An Electron revolves around the nucleus without any orbits.
3. Since most of the particles passed through the foil undeflected it means that most of the volume occupied by an atom is empty.
4. An Atom as a whole is neutral.
5. The deflection of few particles on the foil suggested that there is center of positive particles in an atom called the nucleus of the atom.
6. The complete rebounce of few particles on the gold foil suggested that the nucleus is very dense and hard.

6 0

4 years ago

Use Avogadro's number, 6.02E23, to calculate the number

Zepler [3.9K]

Answer:

<h2>2.408 × 10²¹ is the correct answer!!</h2>

8 0

3 years ago

Predict the products of this reaction: H2 + Cl2 →

Readme [11.4K]

Answer:

2HCl is the product of this reaction 2 is added in order to balance the reaction

8 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

3 years ago