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natta225 [31]
4 years ago
11

At which temperature will the least amount of time pass before someone detects the odor of a fragrant flower?

Chemistry
1 answer:
alukav5142 [94]4 years ago
7 0
<span>0 °C. 8 °C. 29 °C. 15 °C.</span>
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Al comenzar la reacción: N2(g) + 2O2(g) ------&gt; 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al
miss Akunina [59]

Answer:

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Explanation:

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Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al

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3 years ago
Choose the solvent below that would show the greatest freezing point lowering when used to make a 0.20 m nonelectrolyte solution
tatiyna

Answer : Carbon tetrachloride, k_f=29.9^oC/m will show the greatest freezing point lowering.

Explanation :

For non-electrolyte solution, the formula used for lowering in freezing point is,

\Delta T_f=k_f\times m

where,

\Delta T_f = lowering in freezing point

k_f = molal depression constant

m = molality

As per question, the molality is same for all the non-electrolyte solution. So, the lowering in freezing point is depend on the k_f only.

That means the higher the value of k_f, the higher will be the freezing point lowering.

From the given non-electrolyte solutions, the value of k_f of carbon tetrachloride is higher than the other solutions.

Therefore, Carbon tetrachloride, k_f=29.9^oC/m will show the greatest freezing point lowering.

4 0
3 years ago
Write the balanced molecular equation for the neutralization reaction between hcl and ba(oh)2 in aqueous solution. include physi
mr_godi [17]
Each Ba(OH)2 requires 2 Cl to produce one BaCl2 molecule. 

<span>The 2 H's from the 2 HCl molecules combine with the 2 (OH) to form 2 H2O molecules. </span>

<span>2 HCl (aq) + Ba(OH)2 (s) ==> BaCl2 (aq) + 2 H2O (l)

I hope the answer will help you. </span>
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3 years ago
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Answer:

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The volume of a gas at 30◦C and 0.13 atm is 61 mL. What volume will the same gas sample occupy at standard conditions?
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 <span>n= PV/RT </span>
<span>n= 0.44x0.025/0.082x297.5 </span>
<span>n=0.000450913 </span>

<span>22.4L/1 mole = y L/ 0.000450913 </span>
<span>y=0.0101043 L </span>

<span>y = 10.1 mL</span>
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3 years ago
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