1. List the seasons from longest to shortest, including the number of days in each season.

The absorbance of a chlorophyll a standard solution in ethanol (density = 0.785 g/mL) is measured in

Let A= ebc 
 0.765 = 86,300*1.3*c 

Solve for c = approximately 7E-6 Molar
= mols/L soln. 
g = 7E-6*893.49
= about 0.006 g chlorophyll/L soln. 

1000 x 0.785 = 785 g ethanol. 
Conc. = about 0.006g chlorophyll/785 g soln. 

Change that to ppm. by using formula:
(0.006/785)*1E6

5 0

3 years ago

Read 2 more answers

Explain why a random copolymer is obtained when 3,3-dimethyl-1-butene undergoes cationic polymerization.

dezoksy [38]

Solution :

In the field of chemistry, the cationic polymerization is a kind of the chain growth polymerization where the cation initiator transfers the charge to the monomer and makes it more reactive. This kind of polymerization reaction is very sensitive to the temperature. With increase in temperature, the molecular weight as well as the reaction rate decreases rapidly.

Thus in the cationic polymerization of the 3,3-dimethyl-1-butene, the carbonation intermediate is formed and it rearranges itself. The attack o the 2nd alkene ca take place both the carbonations, so that random copolymer is formed.

8 0

3 years ago

Hormones taken as medicine can harm fish when the hormones end up in waterways. What does this show about the impact of chemical

iragen [17]

The answer to this is C.

3 0

3 years ago

How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina

liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 = [A⁻] / [HA] (1)

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 (2)

Replacing (1) in (2):

0.3373 = 0.08255 - [HA] / [HA]

0.3373[HA] = 0.08255 - [HA]

1.3373[HA] = 0.08255

[HA] = 0.06173 moles

Thus:

[A⁻] = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

3 0

3 years ago

A student obtains the following data: Mass of empty, dry graduated cylinder: 21.577 g Volume added of NaCl solution: 4.602 mL Ma

klasskru [66]

Answer: The density of NaCl solution is 3.930 g/mL

Explanation:

We are given:

Mass of cylinder, $m_1$ = 21.577 g

Mass of NaCl and cylinder combined, M = 39.664 g

Mass of NaCl, $m_2$ = $(M-m_1)=(39.664-21.577)g=18.087g$

To calculate density of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Mass of NaCl = 18.087 g

Volume of NaCl solution = 4.602 mL

Putting values in above equation, we get:

$\text{Density of NaCl}=\frac{18.087g}{4.602mL}\\\\\text{Density of NaCl}=3.930g/mL$

Hence, the density of NaCl solution is 3.930 g/mL

7 0

2 years ago