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3 years ago

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Calculate the energy (in joules) of each photon that is emitted from a patch of ground that is at 32 ∘C. Use the λmax of that te

mperature as your wavelength. Note that you can manually enter temperature values in the simulation in the white box.

1 answer:

AlekseyPX3 years ago

3 0

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hope that helps!

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Which functional group is found in methanol?

denis23 [38]

Yes , CH3OH - this is methanol

8 0

3 years ago

Read 2 more answers

frez [133]

Answer:

6.25%

Explanation:

Given data:

Half life of lutetium-117 = 6.75 days

Percentage remaining after 27 days = ?

Solution;

Number of half lives = Time elapsed / half life

Number of half lives = 27 days / 6.75 days

Number of half lives = 4

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

At 3rd half life = 25%/2 = 12.5%

At 4th half life = 12.5%/2 = 6.25%

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3 years ago

What ions are released when water is mixed with a base?

ZanzabumX [31]

Answer:

When dissolved in water, acids donate hydrogen ions (H+). Hydrogen ions are hydrogen atoms that have lost an electron and now have just a proton, giving them a positive electrical charge. Bases, on the other hand, mixed with water yield hydroxide ions (OH-).

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3 years ago

Write a balanced equation for the following neutralization reaction HBr+LiOH<->

Sophie [7]

Answer:

HBr(aq) + LiOH(aq) = LiBr(aq) + H2O(l)

Explanation:

For this reaction, the reactants are the hydrobomic acid and the lithium hydroxide which produces the products lithium bromide and water.

8 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago