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Alik [6]
4 years ago
11

A certain element has 12 neutrons and 12 protons. Assume that this atom has an

Chemistry
1 answer:
KATRIN_1 [288]4 years ago
5 0
The element is magnesium
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What is the ranking order of the degree of polarity? Of HF,HCI,HBr and HI
Savatey [412]

Answer:

The ranking order of the degree of polarity is HF>HCl>HBr>HI

Explanation:

The degree of polarity of a compound depends on the nature electronegative atom that it contains.

      Among Fluorine,chlorine,bromine and iodine the electronegativity of fluorine is highest among all the mentioned atoms.

     Due to high electronegativity HF is the most polar than HCl,HBr and HI and the HI have least polarity as its atomic structure is large and electronegativity is lower than fluorine,chlorine and bromine.

7 0
4 years ago
Please help I will mark as Brainly
Elina [12.6K]
The answer is D because when a scientist publishes a book it helps other scientists and other scientists can help the scientists that publish the book.
7 0
3 years ago
A chemistry student needs 75.0ml of pentane for an experiment. by consulting the crc handbook of chemistry and physics, the stud
vagabundo [1.1K]
Mass=density x volume
0.626 x 75
46.95
3 0
3 years ago
Aspirin is not as____<br> in water as sugar is.
Gre4nikov [31]

Answer:Soluble

Explanation:Soluble:(of a substance) able to be dissolved, especially in water.

6 0
3 years ago
The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the
Sonbull [250]

Answer:

The rate of consumption of NH_{3} is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

where -\frac{\Delta [NH_{3}]}{\Delta t} represents rate of consumption of NH_{3}, -\frac{\Delta [O_{2}]}{\Delta t} represents rate of consumption of O_{2}, \frac{\Delta [N_{2}]}{\Delta t} represents rate of formation of N_{2} and \frac{\Delta [H_{2}O]}{\Delta t} represents rate of formation of H_{2}O.

Here rate of formation of H_{2}O is 3.0 mol/(L.s)

From the above equation we can write-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Here \frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))

So, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Hence, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)  

6 0
3 years ago
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