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8090 [49]
3 years ago
12

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.

Chemistry
1 answer:
AlekseyPX3 years ago
3 0

Attached response


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
Help please!!
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%

7 0
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-).

5 0
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
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