With a mole of skittles placed side by side, you can go around the earth <span>191.17 trillion times.
I hope this helps! If it didn't please reply. Please mark it the brainliest if it did!
</span>(6.022x10^23) x .5" = 3.011x10^23 inches.
<span>3.011x10^23 / 12 in per ft(=.251)
</span><span>.251 /5280 ft per mile = 4.75x10^18 miles
</span><span>4.75x10^18 miles / 24850 mile around the earth = 1.9117x10^14 time around the earth.
</span><span>191.17 trillion times around the earth</span>
Answer:
Rate = −1/2 Δ[N2O] / Δt = 1/2 Δ[N2]/ Δt = Δ[O2]/Δt
Explanation:
Based on the reaction:
2 N2O(g) → 2 N2(g)+O2(g)
The rate of reaction is the change in concentration per mole of reaction.
In this problem, 2 moles of N₂O are consumed, the rate per mole of reaction is:
Rate = -1/2 Δ[N₂O] / Δt
<em>Negative because this reactant is consumed</em>
In therms of N₂:
Rate = 1/2 Δ[N₂] / Δt
<em>1/2 because 2 moles are produced per mole of reaction</em>
And for O₂:
Rate = Δ[O₂] / Δt
Right solution is:
<h3>Rate = −1/2 Δ[N2O] / Δt = 1/2 Δ[N2]/ Δt = Δ[O2]/Δt</h3>
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Answer:
u can eaisly do this just make sure u empty the filter and let dry to prevent mold
Explanation:
Answer:
15.2L at STP
Explanation:
Given reaction expression;
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃ = 0.68mol
Unknown:
Volume of CO₂ produced at STP = ?
Solution:
To solve this problem, we must first find the number of moles CO₂ produced,
1 mole of CaCO₃ will produce 1 mole of CO₂
0.68mole of CaCO₃ will produce 0.68mole of CO₂ at STP
Now;
1 mole of gas occupies a volume of 22.4L at STP;
0.68mole of CO₂ will then occupy 0.68 x 22.4 = 15.2L at STP
<span>The mass and volume of each sample differ from the mass and volume of the other samples. Is it possible for each sample to contain 1 mol of each substance?
</span><span>C) Yes, because the number of moles is not dependent on the mass or the volume.</span>
