Ask question

Ask question

All categories

2 years ago

10

Leslie wants to know which advertising medium works best to promote her electronics store. She surveys her customers over a week

end and creates a bar graph.
Based on her survey, which advertising mediums will work best to publicize Leslie’s in-store specials?

newspaper and printed flyers
radio and TV
radio and Internet
printed flyers and Internet

2 answers:

ryzh [129]2 years ago

7 0

The answer is D - printed flyers and internet

kvasek [131]2 years ago

4 0

Answer:

D

Explanation:

You might be interested in

Fe3+(aq)+6H2O(l)⇌Fe(H2O)63+(aq) : F e 3 + ( a q ) + 6 H 2 O ( l ) ⇌ F e ( H 2 O ) 6 3 + ( a q ) : blank is the Lewis acid and bl

Tasya [4]

Answer:

Lewis acid- Fe3+

Lewis base- water molecule

Explanation:

Acids and bases have been defined in diverse ways. There have been definitions put forward by Arrhenius, Brownstead and Lowry as well as Lewis. Each definition his useful in its own way.

Lewis acids are lone pair acceptors such as metal ions. This implies that in the particular instance of this reaction, Fe3+ is the lewis acid.

Similarly, a Lewis base is a lone pair donor, all ligands are lone pair donors since they donate one or more lone pairs of electrons to Lewis acids. In the particular instance of this reaction, the Lewis base is the water molecule.

6 0

2 years ago

What should be done if particles of precipitate appear in the filtrate?

Volgvan

<span>The instructor should be questioned to see if the filtrate is able to be recycled. This precipitate can contaminate the filtrate, rendering it useless for repeated experiments. If it is able to be recycled, a second pass through the filter might be required to remove the precipitate.</span>

4 0

2 years ago

Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0

ExtremeBDS [4]

Answer:

The freezing point will be $2.9^{0}C$

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

$Depressioninfreezing point=K_{f}Xmolality$

Where

Kf= $20\frac{^{0}C}{m}$

the molality is calculated as:

$molality=\frac{moles_{solute}}{mass_{solvent}}$

$moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol$

$massofcyclohexane=25g=0.025Kg$

$molarity=\frac{0.0045}{0.025}=0.18m$

Depression in freezing point = $20X0.18=3.6^{0}C$

The new freezing point = $6.5^{0}C-3.6^{0}C=2.9^{0}C$

5 0

2 years ago

water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water

Gemiola [76]

Answer is: glycerol because it is more viscous and has a larger molar mass.

Viscosity depends on intermolecular interactions.

The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0

2 years ago

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

2 years ago