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zhannawk [14.2K]
3 years ago
6

The percent abundance of each isotope tells you how many of each kind of isotope exist in every 100 particles. What does relativ

e abundance tell you?
Chemistry
1 answer:
fiasKO [112]3 years ago
3 0

Answer:

The answer is Relative plenitude alludes to the amount of a specific isotope is available in a given measure of test.  

Explanation:

The 'relative plenitude' of an isotope implies the level of that specific isotope that happens in nature. Most components are comprised of a blend of isotopes. The total of the rates of the particular isotopes must indicate 100%. The relative nuclear mass is the weighted normal of the isotopic masses. The percent plenitude of every sort of sweets reveals to you what number of every sort of Aufbau there are in each 100 CANDIES. Percent wealth is additionally relative plenitude. This is only a method for giving us a photo on which kind exists all the more every now and again.

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The freezing point of a solution the freezing point of the pure solvent
babymother [125]

Answer:

This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur.

Explanation:

5 0
2 years ago
Read 2 more answers
LIMITING REACTANT!! Please help I’m very confused.
alexira [117]

Answer:

We'll have 1 mol Al2O3 and 3 moles H2

Explanation:

Step 1: data given

Numer of moles of aluminium = 2 moles

Number of moles of H2O = 6 moles

Step 2: The balanced equation

2Al + 3H2O → Al2O3 + 3H2

Step 3: Calculate the limiting reactant

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

Aluminium is the limiting reactant. It will completely be consumed (2 moles).

H2O is in excess. There will react 3/2 * 2 = 3 moles

There will remain 6 - 3 = 3 moles

Step 4: Calculate moles products

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

For 2 moles Al we'll have 2/1 = 1 mol Al2O3

For 2 moles Al We'll have 3/2 * 2 = 3 moles H2

We'll have 1 mol Al2O3 and 3 moles H2

8 0
3 years ago
A block is found to have volume of 35.3 cm3. It’s mass is 31.7g. Calculate the density of the block
11111nata11111 [884]

Answer:

<h2>1.11 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{35.3}{31.7}  \\  = 1.113564

We have the final answer as

<h3>1.11 g/mL</h3>

Hope this helps you

6 0
2 years ago
Which of these expressions are correct variations of the Combined Gas Law?
mafiozo [28]

Answer:

Both

Explanation:

The combined gas law is also known as the general gas law.

From the ideal gas law we assume that n = 1;

So;

              PV  = nRT

 and then;

                  \frac{P_{1}V_{1}  }{T_{1} }  = \frac{P_{2}V_{2}  }{T_{2} }

   If we cross multiply;

                P₁V₁T₂   = P₂V₂T₁

  So;

         T₁ = T_{2} \frac{P_{1}V_{1}  }{P_{2} V_{2} }

Also;

         V₂  = V_{1} \frac{P_{1} T_{2} }{P_{2} T_{1} }

So from the choices both are correct

3 0
3 years ago
2) A common "rule of thumb" -- for many reactions around room temperature is that the
babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

$\Rightarrow \frac{K_1}{K_2}=  e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

           R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0
2 years ago
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