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ale4655 [162]
2 years ago
5

Which one of the following compounds produces a basic solution when

Chemistry
1 answer:
JulsSmile [24]2 years ago
6 0

Answer:

Rb2O

Explanation:

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vesna_86 [32]

Answer:

bebe bebe bebe

Explanation:

5 0
3 years ago
Using molecular orbital theory, explain why the removal of one electron in O2 strengthens bonding, while the removal of one elec
Nuetrik [128]
First you have a knowledge of bond order which is
 B.O=(no. of electrons in bonding orbital - no. of electrons in non-bonding orbital)÷2  
Note:
bond strength is directly proportional to bond order.
For oxygen:
B.O=(6-2)/2= 2; after the removal of two electrons(removal occur from non-bonding orbital)
B.O=(6-0)/2= 3 (As B.O increased bond strength increased)
For Nitrogen:
B.O=(6-0)/2= 3; after the removal of two electrons(removal occur from bonding orbital)
B.O=(4-0)/2= 2 (As B.O decreased bond strength decreased)

6 0
3 years ago
A laboratory experiment requires 4.8 l of a 2.5 m solution of sulfuric acid (h2so4), but the only available h2so4 is a 6.0 m sto
Airida [17]
To determine the amount of 6.0 M H2SO4 needed for the preparation, equate the number of moles of the 6.0 M and 2.5 M H2SO4 solution. This is done as follows
 
                                             M1 x V1 = M2 x V2

Substituting the known variables,
 
                                             (6.0 M) x V1 = (2.5 M) x (4.8 L)

Solving for V1 gives an answer of V1 = 2 L. Thus, to prepare the needed solution, dilute 2 L of 6.0 M H2SO4 solution with water until the volume reach 4.8 L. 


5 0
3 years ago
Read 2 more answers
A relatively simple way of estimating profit is to consider the the difference between the cost (the total spent on materials an
Maksim231197 [3]
<span>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 “Calculate the profit from producing 75.00kg of propene oxide” (75.00kg) ($152.44 /kg) = $11,433 that answer rounded off to four sig figs, is $11,430</span>
7 0
3 years ago
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 63. g of octane is mi
Sever21 [200]

Answer:

52.1 g is the maximum mass of CO₂, that can be produced by this combustion

Explanation:

Mass of Octane: 63 g

Mass of O₂: 59.4 g

This is a combustion reaction where the products, are always water and CO₂. We define the equation:

2C₈H₁₈ (l)  +  25O₂(g)  →  16CO₂(g)  +  18H₂O(g)

As we have, both mases of each reactant, we must define which is the limiting reagent. We convert the mass to moles:

63 g. 1mol / 114g = 0.552 moles

59.4 g . 1mol / 32g = 1.85 moles

Certainly, the limiting reagent is the oxygen:

2 moles of octane need 25 moles of O₂ to react

Therefore, 0.552 moles of octane must need (0.552 . 25) /2 = 6.9 moles of O₂ (I do not have enough moles of oxygen, I need 6.9 and I only got 1.85 moles)

When we know the limiting reagent we can do the calculations with the stoichiometry of the reaction:

25 moles of O₂ can produce 16 moles of CO₂

Therefore, 1.85 moles of O₂ may produce (1.85 . 16) /25 = 1.18 moles.

We convert the moles to mass to get the final answer:

1.18 mol . 44 g / 1mol = 52.1 g  

5 0
3 years ago
Read 2 more answers
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