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Andreyy89
3 years ago
9

The atoms of Group B elements _____ electrons when they form ions. options: A.either gain or lose B.gain C.cannot be determined

D.lose
Chemistry
2 answers:
grigory [225]3 years ago
6 0

The answer to the question :

The atoms of Group B elements _____ electrons when they form ions .

is :

D) lose

The atoms of Group B elements lose electrons when they form ions because these elements are metals.

Harrizon [31]3 years ago
6 0
They tend to lose electrons to became positive ions( cation ) .
Hope it help :)
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What do chemists use percent yield calculations for in the real world?
STALIN [3.7K]

Answer:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation.  false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need.  false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are.  true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain  the amount of product you want.

8 0
2 years ago
How are the carbon, oxygen, and nitrogen cycles vital to sustaining life on Earth?. a.. Carbon, oxygen, and nitrogen are vital c
kumpel [21]

The correct answer is (D)


All the above are reasons the carbon , oxygen, and nitrogen cycles are vital to life on earth.


The explanation:


because :


1) Carbon, oxygen, and nitrogen are vital components of life on Earth.


2) The carbon, oxygen, and nitrogen cycles allow vital elements to return to usable form by organisms.


3)The carbon, oxygen, and nitrogen cycles are an important interface between biotic and abiotic factors


4) They are all biogeochemical cycles.


5) They all involve an interaction between living and nonliving elements.

6)They are all part of the Earth system.

5 0
3 years ago
Read 2 more answers
Why aren't descriptive investigations repeatable?
Bad White [126]
In descriptive investigations, we still haven't formed any hypothesis yet so we seek information by asking question.

It's not repeatable because repeating the questions over and over again without any clue about what we want to seek is completely waste of time.

Hope this helps xox :)
8 0
3 years ago
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Place the following in order of increasing molar entropy at 298 K.
Genrish500 [490]

Answer:

SO < CO2 < C3H8

Explanation:

Entropy refers to the degree of disorderliness of a system. The standard molar entropy of a substance refers to the entropy of 1 mole of the substance vunder standard conditions.

The molar entropy depends on the number of microstates in the system which in turn depends on the number of atoms in the molecule.

C3H8 has 11 atoms and hence the highest number of microstates followed by CO2 having three atoms and least of all SO having only two atoms.

3 0
2 years ago
RATE LAW QUESTION !
vivado [14]
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
6 0
3 years ago
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