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3 years ago

Identify the limiting reagent in the reaction mixture shown below (red = a2, blue = b2). the balanced reaction is a2+2b2→2ab2

1 answer:

6 0

With the equation <span>A₂+2B₂→2AB₂ we know that one mole of A₂ is necessary to react with two moles of B₂ and with that, we obtain two moles of the product. ( you can see this by just seeing what numbers are before the letters)

Since it's needed more quantity of B than A, B is the limited reagent.
</span>

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If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?

myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30.17}\ year^{-1}$

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Initial concentration $[A_0]$ = ?

Final concentration $[A_t]$ = 12.5 grams

Applying in the above equation, we get that:-

$12.5\ g=[A_0]e^{-0.02297\times 90.6}$

$[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g$

<u>Mass of original sample = 100 g</u>

8 0

3 years ago

Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

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8 0

1 year ago

Why would Magnesium Phosphate (Mg3(PO4)2) not make an aqueous solution? <br><br> Please help!

netineya [11]

Answer:

Basically, all phosphates except Sodium phosphates, Potassium phosphates and Ammonium phosphates are insoluble in water. That, of course, includes Magnesium phosphate.

Explanation:

Hope this helped!

5 0

3 years ago

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lana66690 [7]

I think it will be D or B but my mine answer is D

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3 years ago

What is the kinetic energy of a 10 kg object that has a velocity of 1.5 m/s

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