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Romashka [77]
3 years ago
7

Blank and blank are examples of solids

Chemistry
1 answer:
Ymorist [56]3 years ago
3 0
A box and a laptop are both examples of solids


hope i helped:)
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Is forms from coal, oil, and gas burning physical or chemical weathering
storchak [24]
The answer is chemical weathering
8 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]
<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
What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula is C2H3
iren [92.7K]
To find the molecular formula from the empirical formula, you need to find a multiple (x) that will give you the molar mass of the compound which in the question is 54 g/mol.

If C₂H₃ is the empirical formula
 molar mass of empirical formula = (12 × 2) + (1 × 3) g/mol
                                                     =  27 g/mol
let x = multiple
let molecular formula = C₂ₓ H₃ₓ

            multiple = molecular mass ÷ empirical mass
                         =  54 g/mol ÷  27 g/mol
                         =  2


If molecular formula = C₂ₓ H₃ₓ

then molecular formula = C₂₍₂₎H₃₍₂₎

                                      = <span>C₄H</span>₆

8 0
3 years ago
Based on the information, what is the balanced chemical equation representing the burning of butane?
kompoz [17]

Answer:

OptionA. 2C4H10 + 13O2 —> 8CO2 + 10H20

Explanation:

Butane burns is air (O2) according to the equation:

C4H10 + O2 —> CO2 + H20

Considering the equation, it is evident that it not balanced. Now let us balance the equation as shown below;

There are a total of 4 carbon atoms on the left and 1 carbon atom on the right. It can be balanced by putting 4 in front of CO2 as shown below:

C4H10 + O2 —> 4CO2 + H20

Next, there are 10 hydrogen atoms on the left and 2 hydrogen atoms on the right. Therefore to balance it, put 5 in front of H2O as shown below:

C4H10 + O2 —> 4CO2 + 5H20

Now, there are a total of 13 oxygen atoms on the right and 2 at the left. To balance it, put 13/2 in front of O2

as shown below

C4H10 + 13/2O2 —> 4CO2 + 5H20

Now we multiply through by 2 clear off the fraction and we obtained:

2C4H10 + 13O2 —> 8CO2 + 10H20

6 0
3 years ago
The atmosphere is about 21% oxygen. At a pressure of 749 torr, what is the partial pressure of O2?
Kitty [74]

Answer:

157.29 Torr

Explanation:

This question may be solved by the mole fraction.

Mole fraction → Moles of gas / Total moles = Partial pressure of gas / Total pressure

If the atmosphere is about 21% you assume that the mole fraction is 0.21 so:

0.21 = Partial pressure of gas / Total pressure

0.21 = Partial pressure of gas / 749 Torr

Partial pressure of the gas = 749 Torr . 0.21 →  157.29 Torr

3 0
3 years ago
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