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Verdich [7]
1 year ago
14

When the temperature of a 3. 0-l sample of a gas is dropped from 200°c to 100°c, what will be the final volume of the gas sample

?.
Chemistry
1 answer:
ipn [44]1 year ago
4 0

P1V1T1=P2V2T2 Add 273 to convert degrees Celsius to Kelvin:

∴200×25/298=250×V2/273, ∴V2=200×25×273/298×250, ∴V2=18.32L

<h3>Where is the volume equation?</h3>

The basic formula for volume is length, breadth, and height, as opposed to length, width, and height for the area of a rectangular shape. The calculation is unaffected by how you refer to the various dimensions; for instance, you can use "depth" instead of "height."

<h3>What is chemistry using volume units?</h3>

Volume, which is measured in cubic units, is the 3-dimensional space occupied by matter or encircled by a surface. The cubic meter (m3), a derived unit, is the SI unit of volume.

to know more about volume and temperature here:

brainly.com/question/12050285

#SPJ4

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How many grams are in 2.3 x 10^ -4 moles of calcium phosphate? ​
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Mass = 713.4 ×10⁻⁴ g

Explanation:

Given data:

Number of moles of calcium phosphate = 2.3×10⁻⁴ mol

Mass of calcium phosphate = ?

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Number of moles = mass/molar mass

Molar mass of calcium phosphate is 310.18 g/mol

by putting values,

2.3×10⁻⁴ mol = mass / 310.18 g/mol

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3 years ago
Massive amounts of energy are contained in an atom's: Select one:
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4 years ago
What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M<br> solution?
love history [14]

Answer:

Approximately 19.1\; \rm g.

Explanation:

<h3>Number of moles of formula units of magnesium sulfate required to make the solution</h3>

The unit of concentration in this question is "\rm M". That's equivalent to "\rm mol \cdot L^{-1}" (moles per liter.) In other words:

c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}.

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L.

Calculate the number of moles of \rm MgSO_4 formula units required to make this solution:

\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}.

<h3>Mass of magnesium sulfate in the solution</h3>

Look up the relative atomic mass data of \rm Mg, \rm S, and \rm O on a modern periodic table:

  • \rm Mg: 24.305.
  • \rm S: \rm 32.06.
  • \rm O: 15.999.

Calculate the formula mass of \rm MgSO_4 using these values:

M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}.

Using this formula mass, calculate the mass of that (approximately) 0.159\; \rm mol of \rm MgSO_4 formula units:

\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}.

Therefore, the mass of \rm MgSO_4 required to make this solution would be approximately 19.1\; \rm g.

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3 years ago
When sodium atoms (Na) and chlorine atoms (CI) join to make
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