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aliina [53]
3 years ago
8

When an atom of n-14 is bombarded by an alpha particle, the single product is?

Chemistry
2 answers:
riadik2000 [5.3K]3 years ago
5 0

<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

Alpha particle is released when an atom undergoes alpha decay. This particle carries a charge of +2 units and has a mass of 4 units. This is also considered as a helium nucleus.

In the question:

An atom of N-14 is bombarded with alpha particle, so the equation for this reaction follows:

_7^{14}\textrm{N}+_2^4\alpha \rightarrow _9^{18}\textrm{F}

Hence, the correct answer is Option c.

aliya0001 [1]3 years ago
4 0
For the answer to the question above asking w<span>hen an atom of n-14 is bombarded by an alpha particle, the single product is?

</span> <span>You're starting with 14/7 N, correct? 

An alpha particle is two protons, two neutrons, which is 4/2, correct? 
</span><span>So I</span> think the answer to your question is the third one which is <span>c. 18/9 f </span>
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For the following example, list the given and unknown information (including gratis or moles)
Setler79 [48]

Answer:

9.6 moles O2

Explanation:

I'll assume it is 345 grams, not gratis, of water.  Hydrogen's molar mass is 1.01, not 101.

The molar mass of water is 18.0 grams/mole.

Therefore:  (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).

The balanced equation states that:  2H20 ⇒ 2H2 +02

It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.

get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2

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2 years ago
Earthquake damage causes two rabbits to be separated from the rest of the rabbits in their large habitat. They have no way to ge
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3 years ago
Describe and compare the building blocks, general structires and biological functions of carbohydrates lipids proteins and nucle
Nata [24]

Answer:

Explanation:

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Monosaccharides which are n>3 (Triose) are the aldose and ketose.

They are the simpleat and smallest form and they are Glucose, fructose and galactose

Disaccharides are structure of the combination of the monosaccharides by glycosidic bond and they are sucrose, lactose, maltose etc

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Lipids(triglycerides) are solid fats or series of repeated fats at room temperature, they are insoluble in water both soluble in some organic solvents. They are also composed of glycerides (3 molecules). Its structure is composed of two parts, the soluble part composing of the -COOH group and the insoluble part that can be saturated or unsaturated hydrocarbon chain

Saturated fats - CH3(CH2)nCOOH

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Secondary structure involves the folding of this chain into alpha helix or beta pleated.

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6 0
4 years ago
How will you separate a solution of benzene and chloroform?
Troyanec [42]

Answer:

\huge\boxed{\sf Distillation}

Explanation:

  • State of benzene at RTP = liquid
  • State of chloroform at RTP = liquid
  • Boiling point of benzene = 80.1 °C
  • Boiling point of chloroform = 61.2 °C

Since, both of the chemicals are liquids, we can separate it by the process of distillation.

<u>Distillation:</u>

  • is the process in which we separate two liquids on the basis of their difference in boiling points.

<u>How it works:</u>

Since chloroform has less boiling point, it will evaporate and collected first and benzene will follow it after sometime.

- Apparatus of distillation is in the attached file.

\rule[225]{225}{2}

7 0
2 years ago
In an aqueous solution of a certain acid the acid is 4.4% dissociated and the pH is 3.03. Calculate the acid dissociation consta
NeX [460]

Answer:

4.1x10⁻⁵

Explanation:

The dissociation of an acid is a reversible reaction, and, because of that, it has an equilibrium constant, Ka. For a generic acid (HA), the dissociation happens by:

HA ⇄ H⁺ + A⁻

So, if x moles of the acid dissociates, x moles of H⁺ and x moles of A⁻ is formed. the percent of dissociation of the acid is:

% = (dissociated/total)*100%

4.4% = (x/[HA])*100%

But x = [A⁻], so:

[A⁻]/[HA] = 0.044

The pH of the acid can be calcualted by the Handersson-Halsebach equation:

pH = pKa + log[A⁻]/[HA]

3.03 = pKa + log 0.044

pKa = 3.03 - log 0.044

pKa = 4.39

pKa = -logKa

logKa = -pKa

Ka = 10^{-pKa}

Ka = 10^{-4.39}

Ka = 4.1x10⁻⁵

4 0
3 years ago
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