At STP, which 2.0-gram sample of matter uniformly fills a 340-milliliter closed container?
Answers
Answer;
D, Xe (g)
Solution and explanation;
If 2g has a volume of 340ml.
Density is 1000/340*2 = 5.88g/litre.
-This rules out the two solids, choices 2) &3)
If 1 litre has mass 5.88g,
then 22.4 liters (volume at STP) has mass 5.88*22.4 = 131.8g/mol
molar mass Br2 = 80*2 = 160g/mol NO
molar mass Xe = 131.3g/mol = YES.
Answer is Xe
Final answer:
To determine the matter filling a closed container at STP given a 2.0 gram sample, you can use the ideal gas law equation to find the number of moles. Then, divide the number of moles by the volume and the gas constant to solve for the pressure. Compare the pressure obtained to known substances' vapor pressures at STP to identify the matter.
Explanation:
The question is asking for the specific type of matter that would uniformly fill a 340-milliliter, closed container at STP (Standard Temperature and Pressure) when given a 2.0-gram sample. To determine the matter, we can use the ideal gas law equation, PV = nRT, and rearrange it to solve for n, the number of moles. Then, we can use the molar mass of the substance to find its identity.
First, convert the volume from milliliters to liters by dividing it by 1000: 340 mL ÷ 1000 = 0.34 L. Next, convert the mass from grams to moles using the molar mass of the substance:
1.(Conversion factor) Given: 2.0 g sample, 1 mole = molar mass
2.(Calculation) Moles of substance = 2.0 g ÷ molar mass
Once you have the number of moles, divide it by the volume (in liters) and the universal gas constant (0.0821 L·atm/mol·K) and solve for the pressure:
1.(Ideal Gas Law) PV = nRT
2.(Substitution) P × 0.34 L = n × 0.0821 L·atm/mol·K
3.(Isolation) P = (n × 0.0821 L·atm/mol·K) ÷ 0.34 L
After solving for the pressure, compare it to known substances' vapor pressures at STP to determine the identity of the matter in the container.
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Ans: A)
HBrO and HBrO₃ are oxyacids where the acidic strength increases with the increase in the number of atoms attached to the central atom.
In both acids, oxygen is the most electronegative atom. In HBrO, the B atom is linked to only one O atom. In contrast, there are 3 electronegative O atoms surrounding the central B atom in HBrO₃ which would make the OH bond more polar and easily accessible. Thus, HBrO₃ tends to lose a proton readily than HBrO making the former more acidic.
HBrO is a weaker acid than HBrO3 because the H-O bond in HBrO is less polar than the H-O bond in HBrO3. In a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s), making the acid stronger.
The acid strength of HBrO is weaker than HBrO3 because the H-O bond in HBrO is less polar than the H-O bond in HBrO3 (Option A). In a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). This stronger attraction of oxygen for the electrons in the O-H bond makes the hydrogen more easily released, resulting in a stronger acid (Option E). Therefore, HBrO3 is a stronger acid than HBrO.
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Answers
There are approximately 4.52 x 10^23 atoms in 0.750 moles of zinc.
How to find the number of atoms
To determine the number of atoms in a given amount of a substance, you can use Avogadro's number, which is approximately 6.022 x 10^23 atoms/mol.
Given that you have 0.750 moles of zinc, you can calculate the number of atoms using the following steps:
Multiply the given number of moles by Avogadro's number:
0.750 moles * (6.022 x 10^23 atoms/mol) = 4.5165 x 10^23 atoms
Round the result to an appropriate number of significant figures:
Since the value given has three significant figures, the final answer should be rounded accordingly. Therefore, the number of atoms in 0.750 moles of zinc is approximately 4.52 x 10^23 atoms.
So, there are approximately 4.52 x 10^23 atoms in 0.750 moles of zinc.
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It is
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Answer:
The rubber becomes brittle and can break in your hand. The explanation for why this happens concerns cross-linking bonds. Ultra-violet light from the sun provides the polymer molecules with the activation energy they need to be able to form more cross-links with other chains.
Final answer:
If excessively excessive cross-links are formed in rubber, it becomes overly rigid and brittle, hampering its natural elasticity and strength. The increase in cross-links restricts the moving of the polymer chains, undermining the effectiveness of the rubber in many applications.
Explanation:
The formation of cross-links in rubber significantly affects its properties. In the case where too many cross-links are formed, the rubber is likely to become overly rigid and brittle. This is because the cross-links restrict the movement of the polymer chains, which reduces flexibility and elasticity. As a result, too many cross-links can compromise the usefulness of rubber for many applications, which require its natural elasticity and strength. For example, in a rubber band, if too many cross-links were formed, then it would be less stretchy and snap more easily when stretched.
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Answer:to show the structures of a plant cell,add a cell wall around the membrane and increase the size of the vaculoe insidethe cell, draw green ovals to represent chloroplasts.
Explanation:
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Answer: Out of all the options presented above the one that cannot be used to define a base is answer choice A) a hydronium ion donor in a reaction. As we already know a base is a compound that is a proton (H¹⁺) acceptor.
I hope it helps, Regards.
I’ll give brainiest if you answer in the next 15 mins
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Final answer:
The magnification of Robert's drawing of a wood louse can be obtained by dividing the size of the drawing (5mm) by the actual size of the wood louse (0.4mm). This gives a magnification of 12.5 times.
Explanation:
The magnification of a drawing can be calculated using the method 'Image size/Actual size'. In this question, Robert's drawing is of a size 5mm while the actual size of the wood louse is 0.4mm. Substituting these values into the formula, you would get 5mm/0.4mm which equates to a magnification of 12.5 times.
The magnification of Robert's drawing of a wood louse can be calculated by dividing the size of the drawing by the actual size, resulting in a magnification of 12.5.
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Answer:
Magnification is the number of times larger an image is compared with the real size of the object.
Magnification = Image / Actual = 5/0.4 =12.5
Explanation: