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777dan777
3 days ago
5

Question 2 Here are four sketches of pure substances. Each sketch is drawn as if a sample of the substance were under a microsco

pe so powerful that individual atoms could be seen. Decide whether each sketch shows a sample of an element, a compound, or a mixture.

Chemistry
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A gas sample of argon, maintained at constant temperature, occupies a volume of 500. l at 4.00 atm. what is the new volume if th
lorasvet [2795]
Boyle's law describes the relationship between gas pressure and volume. 
It asserts that at a constant temperature, pressure is inversely proportional to gas volume.
PV = k
where P represents pressure, V denotes volume, and k is a constant. 
P1V1 = P2V2
where the parameters for the initial condition are on the left, and the parameters for the second condition appear on the right side of the formula.
By substituting values into the equation: 4.00 atm x 500 L = 8.0 atm x V
V calculates to 250 L.
Thus, the new volume becomes 250 L.
6 0
2 months ago
Read 2 more answers
One cubic millimeter (mm3) of blood contains 7.0 x 106 red blood cells. How many red blood cells are in 1.0 L of blood?
Anarel [2989]
The problem provides a conversion factor---> 1 mm3= 7.0 x 10^6 RBC. Therefore, to determine the quantity of red blood cells in your sample, we must first convert Liters to cm3 using the conversion factor--> 1 mL= 1 cm3

I have shared how to resolve this issue.


7 0
1 month ago
1. Use the following thermochemical equation.
KiRa [2933]

Answer:

There's a lot to address, so I'm uncertain if I can tackle this; it feels overwhelming. Perhaps you could simplify it for me as it's quite extensive.

Explanation:

  • I wish I had the answers.
  • It's too complex.
  • This chemistry isn't familiar to me.
  • If you have another chemistry-related question, feel free to ask.
  • This is just too difficult.
7 0
3 months ago
How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?
KiRa [2933]

Response:

To reach the answer, 465.6 mg of MgI₂ is required.

Detailed Explanation:

We need to establish the moles of ion I⁻ in the resulting solution.

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol.

In the initial solution, there was 0.087 M KI, which we can similarly convert into moles, yielding 0.02242 mol.

This indicates we require an additional amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. Since each molecule of MgI₂ produces two I⁻ ions, we divide 0.00335 by 2 to determine the moles of MgI₂, giving us 0.001675 mol.

Consequently, the quantity of MgI₂ to be added is:

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0
2 months ago
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