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olganol [36]
3 years ago
13

A class wants to find the ideal temperature for plants to grow in a terrarium. If the students set up several terrariums, what s

hould they do to get the best results?
a. Plant different kinds of plants in each terrarium.
b. Use a light bulb with different brightness in each terrarium.
c. Add different amounts of water to each terrarium
d. Use different fertilizers in each terrarium


Please help!!
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
3 0

It would be B because you want to know which TEMPERATURE is the best for the plant. Different brightness for a bulb will give different amounts of heat. Since the amount of heat is your INDEPENDENT VARIABLE, you have to change the amount of heat for each plant, so the brightness of bulbs. Therefore, B is your answer!

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A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t
jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x  (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is  initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for  weak acids for the equilibria:

protonated lidocaine + H₂O   ⇆  lidocaine + H₃O⁺

                      protonated lidocaine          lidocaine        H₃O⁺

Initial(M)               0.22                                       0                  0

Change                   -x                                      +x                 +x

Equilibrium          0.22 - x                                  x                    x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] =  x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸  = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22  - x  ≈ 0.22

and solve for x. The pH  will then  be the negative log of this value.

8.71 x 10⁻⁷  = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0
3 years ago
Find the number of Li atoms in 1.50 mole of Li
ale4655 [162]
The molar mass of Li->7g/mol
If 1mol of Li is 7g/mol
1.50mol of Li would be 10.5g/mol
7 0
3 years ago
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Aneli [31]
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4 0
3 years ago
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A mixture of 10.0 g of Ne and 10.0 g Ar have a total pressure of 1.6 atm. What is the partial pressure of Ne?
Natasha2012 [34]
If u add u will get ur answer
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5. 16.3 g of NaCl is dissolved in water to make 1.75 L of solution. What is the molarity of this solution? A 0.159 M B 0.278 M C
ira [324]

Answer: The molarity of this solution is 0.159 M.

Explanation:

Given: Mass of solute = 16.3 g

Volume = 1.75 L

Number of moles is defined as the mass of substance divided by its molar mass.

Hence, moles of NaCl (molar mass = 58.44 g/mol) ar calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{16.3 g}{58.44 g/mol}\\= 0.278 mol

Molarity is the number of moles of a substance present in a liter of solution.

So, molarity of the given solution is calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.278 mol}{1.75}\\= 0.159 M

Thus, we can conclude that the molarity of this solution is 0.159 M.

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