The mass of air in room as per given density is 53.2 kg.
To calculate the mass of air contained in a room, we can use the formula:
mass = density x volume
Here, the given density of air is 1.29 g/dm³ at 25°C. We can convert the dimensions of the room to decimeters (dm) by multiplying by 10:
Length = 2.50 m × 10 = 25 dm
Width = 5.50 m × 10 = 55 dm
Height = 3.00 m × 10 = 30 dm
Now, we can calculate the volume of the room by multiplying the three dimensions:
Volume = length x width x height
Volume = 25 dm x 55 dm x 30 dm
Volume = 41,250 dm³
Finally, we can use the formula to calculate the mass of air:
mass = density x volume
mass = 1.29 g/dm³ x 41,250 dm³
mass = 53,212.5 g or 53.2 kg
Therefore, the mass of air contained in the room is approximately 53.2 kg.
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Explanation:
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largest atomic radii.
Answer:
left to right across a period when it decreases and when it increases top to bottom in a group,
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The observation in this instance relates to the quantity of heat needed to melt ice, and it is expressed in terms of weights (2 lbs and 1 lb) and a comparison (twice the amount).
Without going into detail into the different molecules or their interactions, it concentrates on the general behaviour and characteristics of the substance (ice) as a whole.
A microscopic description, on the other hand, would describe the behaviour in terms of the molecular or atomic interactions that take place at the particle level. It would go into ideas such as the amount of heat required to dissolve the intermolecular interactions between water molecules.
Therefore, the observation regarding how much heat is needed to melt ice is a macroscopic description since it ignores the underlying molecular interactions in favour of the substance's general behaviour and qualities.
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The observation that melting 2 lbs of ice requires twice the heat of melting 1 lb is a macroscopic description, focusing on observable properties and behavior without exploring microscopic details.
This observation is a macroscopic description of chemical behavior. Macroscopic descriptions involve the properties and behavior of substances on a large scale that can be observed directly, without delving into the molecular or atomic details. In this case, the statement refers to the amount of heat required to melt a certain quantity of ice, and it is expressed in terms of macroscopic, measurable quantities (pounds of ice and the associated heat).
The macroscopic observation does not provide insight into the molecular or atomic interactions within the ice but rather focuses on the overall behavior of the substance. The concept that the amount of heat required to melt 2 lbs of ice is twice that needed for 1 lb of ice is a statement about the material's behavior at a larger scale.
This observation aligns with the macroscopic principles of heat and phase transitions, where the heat required for a phase change is directly proportional to the mass of the substance undergoing the transition. The macroscopic perspective is concerned with observable properties and measurements, making it a practical and accessible way to describe chemical behavior without delving into microscopic details.
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Answer:
To determine the value of Kp for the given equilibrium, we need to use the partial pressures of the gases involved.
In the balanced equation: 2 HI (g) ⇌ H₂ (g) + I₂ (g), the stoichiometric coefficients are 2, 1, and 1 respectively.
At equilibrium, the expression for Kp is given by:
Kp = (P(H₂) * P(I₂)) / (P(HI)²)
Using the provided partial pressures:
P(HI) = 1.9 atm
P(H₂) = 7.9 atm
P(I₂) = 2.3 atm
Substituting these values into the expression for Kp:
Kp = (7.9 * 2.3) / (1.9²)
Kp ≈ 19.5 / 3.61
Calculating the result:
Kp ≈ 5.4
Therefore, the value of Kp for the given equilibrium is approximately 5.4.
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Answer:
4.93g are extracted
Explanation:
Partition coefficient (P) is defined as the ratio of solute dissolved in the organic solvent and the solute dissolved in the aqueous phase.
That is:
P = 7.5 = Concentration in dichloromethane / Concentration in water.
Knowing this, in the first extraction with 25mL of dichloromethane you will extract:
7.5 = (X/25mL) / (5g - X) / 100mL
Where X is the amount of compound A that is extracted.
7.5 = 100X / (125 - 25X)
937.5 - 187.5X = 100X
937.5 = 287.5X
3.26g of A are extracted in the first extraction.
In water will remain 5g - 3.26g = 1.74g
In the second extraction you will extract:
7.5 = (X/25mL) / (1.74g - X) / 100mL
7.5 = 100X / (43.5 - 25X)
326.25 - 187.5X = 100X
326.25 = 287.5X
1.13g are extracted in the second extraction.
And remain: 1.74g - 1.13g = 0.61g
In the third extraction you will extract:
7.5 = (X/25mL) / (0.61g - X) / 100mL
7.5 = 100X / (15.25 - 25X)
114.375 - 187.5X = 100X
114.375 = 287.5X
0.40g are extracted in the third extraction.
And remain: 0.61g - 0.40g = 0.21g
In the second extraction you will extract:
7.5 = (X/25mL) / (0.21g - X) / 100mL
7.5 = 100X / (5.25 - 25X)
39.375 - 187.5X = 100X
39.375 = 287.5X
0.14g are extracted in the fourth extraction.
Thus, after the three extractions you will extract: 0.14g + 0.40g + 1.13g + 3.26g = 4.93g are extracted
The process involves using the partitioncoefficient to determine how much of Compound A will prefer the dichloromethane solvent over the water. Following a calculation process through four rounds of extraction, it is concluded that approximately 4.999g of Compound A will be extracted using four 25mL portions of dichloromethane.
The partition coefficient of a compound is a measure of how much it prefers one solvent over another. Given that the partition coefficient of Compound A is 7.5 in dichloromethane with respect to water, we can predict how much of this compound could be extracted using four separate 25 mL portions of dichloromethane.
Here's the step-by-step calculation process:
In total, around 4.999g of compound A will be extracted using four 25mL portions of dichloromethane.
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Answer:
Photosynthesis
Explanation:
Question:
Answer:
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Explanation:
Ionic equation
NaCl(aq) --> Na+(aq) + Cl-(aq)
Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)
In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.
Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)
= 142 g/mol
Molecular weight of NaCl = 23 + 35.5
= 58.5 g/mol
Masses
% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100
= 46/142 * 100
= 32.4%
% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100
= 23/58.5 * 100
= 39.3%
Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.
You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.
No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.
For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.
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