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mario62 [17]
3 years ago
5

At which temperature would air hold the LEAST water vapor?

Chemistry
1 answer:
AleksAgata [21]3 years ago
4 0

Answer:

2°C

Explanation:

The answer is D because cooler air holds less water vapor than warmer air, and 2°C is the coldest choice, so there's your answer!

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How do you determine the charges on the ions in an ionic compound?
Hitman42 [59]

Answer:

To find the ionic charge of an element you'll need to consult your Periodic Table. On the Periodic Table metals (found on the left of the table) will be positive. Non-metals (found on the right) will be negative.

Explanation:

3 0
3 years ago
I need help
liubo4ka [24]

The activation energy in the diagram is 43.8 kcal/ mole, letter C. You have to note that activation energy is the energy needed for the reaction to occur and produce products. Therefore, the spike after H2 and I2 is reacted is the activation energy of the reaction.

3 0
4 years ago
Assertion (A): An orbital cannot have more than two electrons, moreover, if an orbital has two electrons they must have opposite
ArbitrLikvidat [17]

Answer: Its option "A" Both A and R are true and R is the correct explanation of A.

Hope it helps

5 0
3 years ago
Read 2 more answers
1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%
bagirrra123 [75]

Answer:

m_{PbI_2}=18.2gKI

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2

Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:

n_{Pb(NO_3)_2}=96.7mL*\frac{1.134g}{mL}*\frac{0.14gPb(NO_3)_2}{1g}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  =0.0464molPb(NO_3)_2\\\\n_{KI}=99.8mL*\frac{1.093g}{mL}*\frac{0.12gKI}{1g}*\frac{1molKI}{166gKI}  =0.0789molKI

Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):

n_{KI}^{consumed\ by\ Pb(NO_3)_2}=0.0464molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0928molKI

Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:

m_{PbI_2}=0.0789molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gKI

Best regards.

3 0
4 years ago
Assume that you did the following dilutions
MrMuchimi
Below is the solution. I hope it helps. 

CfVf = CiVi 
<span>Cf = (CiVi)/ Vf </span>

<span>i. Cf = [ (10^-6 mol / L) (1 mL) (1L / 1000 mL) ] / [ (1kL) (1000L / 1 kL) ] = 1x10^-12 M → use as Ci in next dilution </span>
<span>ii. Cf = 1x10^-19 M → use as Ci in next dilution </span>
<span>iii. Cf = 1x10^-22 M </span>
6 0
3 years ago
Read 2 more answers
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