When preparing dilute solutions of an acid, carefully pour -

Each orbit surrounding an atom is allowed _____.

Juli2301 [7.4K]

Two electrons is your answer glad to help!

4 0

2 years ago

In the following list, only ________ is not an example of a chemical reaction. the production of hydrogen gas from water the tar

Vika [28.1K]

Answer: Option (c) is the correct answer.

Explanation:

A chemical reaction is defined as the reaction where a chemical bond will break in order to form a new bond due to the formation of a new substance.

For example, $2Na + Cl_{2} \rightarrow 2NaCl$

Here, NaCl is the new substance that is formed. A chemical reaction will always bring change in chemical composition of a substance.

The production of hydrogen gas from water, the tarnishing of a copper penny, charging a cellular phone and burning a plastic water bottle are all chemical reactions.

Whereas a reaction where no change in chemical composition of a substance takes place is known as a physical reaction.

For example, chopping a log into sawdust will change the shape but it will not bring any change in chemical composition of the substance.

Thus, we can conclude that in the following list, only chopping a log into sawdust is not an example of a chemical reaction.

8 0

2 years ago

If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln

Ksju [112]

Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.

5 0

2 years ago

Read 2 more answers

Element X is a radioactive isotope such that every 82 years, its mass decreases by half. Given that the initial mass of a sample

lesantik [10]

Answer: 17 years

Explanation:

Expression for rate law for first order kinetics for radioactive substance is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{82years}=8.4\times 10^{-3}years^{-1}$

b) for 8900 g of the mass of the sample to reach 7700 grams

$t=\frac{2.303}{8.4\times 10^{-3}}\log\frac{8900}{7700}$

$t=17years$

Thus it will take 17 years

4 0

2 years ago

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan

Advocard [28]

Answer:

Manganese, Fifth transition element

[X] 3d⁶ 4s¹

Iron, Sixth transition element

[X] 3d⁶ 4s²

Explanation:

Complete Question

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works as it does in our universe, except that there are six d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the sixth and seventh elements in the first transition series in Universe L. Note; you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series.

Solution

In our universe, there are 5 d orbitals.

And according to Aufbau's principles that electrons fill the lower energy orbitals before they fill higher energy orbitals and Hund's Rule that states that electrons are fed singly to all the orbitals of a subshell before pairing occurs.

The fifth and sixth transition elements in our universe is then Manganese and Iron respectively.

Manganese - [Ar] 3d⁵ 4s²

Iron - [Ar] 3d⁶ 4s²

So, in the new universe L, where there are six d orbitals, for manganese, the fifth transition metal, because half filled orbitals are more stable than partially filled orbitals (that woukd have been rhe case if we leave 5 electrons on the 3d orbital), the 4s orbital is filled to half of its capacity and the one electron removed from the 4s is used to fill the six 3d orbital to half of its capacity too.

For the sixth transition element, the new extra electron just fills the lower energy 4s orbital, leaving the six 3d orbitals all half-filled.

Hence, they both have ground state configurations of

- Manganese, Fifth transition element

[X] 3d⁶ 4s¹

- Iron, Sixth transition element

[X] 3d⁶ 4s²

Hope this Helps!!!

7 0

2 years ago