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3 years ago

For a reaction 2A + B 2C, with the rate equation: Rate = k[A]2[B]

1 answer:

3 0

For reaction

2 A + B ------------> 2 C

Rate = K [ A ]² [ B ]

 the order with respect to A is 2 and the order overall is 3.

hope this helps!

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PLEASE HELP THEY ARE EASY QUESTIONS THERE ARE JUST A LOT Please answer these. The tables needed for question 7 are in the pictur

Andrew [12]

Answer:

Explanation:

2020

4 0

2 years ago

C. Calculate the number of moles in 62g of CO2

worty [1.4K]

Answer:

32÷5

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4 0

2 years ago

Read 2 more answers

Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be

vlabodo [156]

Answer:

The theoretical yield of alum is 19.873 grams and percent yield for this alum synthesis is 45.33%.

Explanation:

Mass of bottle = 10.221 g

Mass of bottle with aluminium pieces = 11.353 g

Mass of aluminium = 11.353 g - 10.221 g = 1.132 g

Mass of alum and bottle = 19.230 g

Mass of alum = 19.230 g - 10.221 g = 9.009 g

Experimental yield of alum = 9.009 g

Theoretical yield of alum:

$2Al(s) +2 KOH(aq) +4H_2SO_4(aq)+10 H_2O(l) \rightarrow 2 KAl(SO_4)_2.12 H_2O(s)+3H_2(g)$

Moles of aluminium = $\frac{1.132 g}{27 g/mol}=0.041926 mol$

According to reaction, 2 moles of aluminum gives 2 moles of alum.

Then 0.041926 mol aluminium will give :

$\frac{2}{2}\times 0.041926 mol=0.041926 mol$ of alum.

Mass of 0.041926 moles of alum:

0.041926 mol × 474 g/mol= 19.873 g

Theoretical yield of alum = 19.873 g

Percentage yield:

$\% Yield =\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the alum:

$\% Yield =\frac{ 9.009 g}{19.873 g}\times 100=45.33\%$

The theoretical yield of alum is 19.873 grams and percent yield for this alum synthesis is 45.33%.

7 0

3 years ago

A relatively simple way of estimating profit is to consider the the difference between the cost (the total spent on materials an

Maksim231197 [3]

Answer: it shows that 1mol mCPHA provides the oxygens to 1 mol of propene, to make 1 mole of C3H6O so: 1 mol C3H6 & 1 mol mCPHA --> 1 mol C3H6O using molar masses, that equation becomes: 42.08grams C3H6 & 172.57grams mCPHA --> 58.08grams C3H6O which is: 42.08 kg C3H6 & 172.57 kg mCPHA --> 58.08 kg C3H6O to produce 1 kg of C3H6O, this becomes: 42.08 / 58.08 kg C3H6 & 172.57 / 58.08 kg mCPHA --> 58.08 /58.08 kg C3H6O which is: 0.72452 kg C3H6 & 2.9712 kg mCPHA --> 1 kg C3H6O but because the reaction gives only a 96% yield, we scale up the reactants to get that desired 1 kg of C3H6O (0.72452 kg ) (100/96) C3H6 & (2.9712 kg) (100/96) mCPHA --> 1 kg C3H6O which is: 0.75471 kg C3H6 & 3.095 kg mCPHA --> 1 kg C3H6O ========= costs per kg of C3H6O produced: (0.75471 kg C3H6) ($10.97 per kg) = $8.279 (3.095 kg mCPHA) ($5.28 per kg) = $16.342 & (0.75471 kg C3H6) / (0.0210 kg C3H6 / L dichloromethane) = 35.939 Litres dichloromethane (35.939 Litres dichloromethane) ($2.12 per L) = $ 76.19 & waste disposal is $5.00 per kilogram of propene oxide produced total cost, disregarding labor,energy, & facility costs: $8.279 & $16.342 & $ $ 76.19 & $5.00 = $105.81 per kg C3H6O produced ========== profit: ($258.25 / kg C3H6O) - ($105.81 cost per kg) = $152.44 profit /kg â€śCalculate the profit from producing 75.00kg of propene oxideâ€ť (75.00kg) ($152.44 /kg) = $11,433 that answer rounded off to four sig figs, is $11,430

7 0

3 years ago

Is anyone good at chemistry if so can someone help me please ?

saveliy_v [14]

We're given the [OH⁻] as 8.34 × 10⁻¹² M. Using the formula pOH = -log[OH⁻], the pOH of this solution would be -log(8.34 × 10⁻¹²) ≈ 11.08.

The pOH is, for lack of a better term, the "opposite" of pH: A pOH of 7 is neutral; a pOH less than 7 is basic; and a pOH greater than 7 is acidic.

This follows from the relation, pH + pOH = 14. In this case, with a pOH of 11.08, our pH would be 14 - 11.08 = 2.92, which is acidic (pH < 7).

Thus, the correct answer choice is B.

4 0

3 years ago