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Stella [2.4K]
3 years ago
13

Calculate the root mean square velocity of gaseous argon atoms at 26 ∘c.

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
5 0
By substitution in this formula :
V= √3RTM
when R = 8.31451 J.mol^-1K^-1
T is the tempreture in K = 26 °C + 273 = 299 k
M is the molar mass of argon Kg= 39.9 g = 39.9/1000= 0.0399 Kg
∴V = √3*8.31451 * 299 / 0.0399 
∴ V = 432 m/s 
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In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 53.88 g of MgI₂ form, what is the pe
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We know the law of conservation of mass

  • It states that mass is neither formed nor destroyed in any chemical reaction.
  • Mass of reactants=Mass of products.

Here

  • Mg and I_2 are reactants
  • MgI_2 is product with some yield.
  • Mass of reactants=10+60.0=70.0g
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  • Mass of yield=Product-MgI_2=70-53.88=16.12g

Lets find the percentage

\\ \tt\hookrightarrow \dfrac{Mass\:of\:yield}{Total\:mass}\times 100

\\ \tt\hookrightarrow \dfrac{16.12}{70}\times 100

\\ \tt\hookrightarrow 0.23028(010)=23.028\%

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