Answer:
influencing consumer tastes
increasing product awareness
promoting company branding
Explanation:
Advertising is basically a form of communication using creative ideas and communicating benefits of the products. Advertising plays a very crucial role in product business and some of the important uses of advertising are as follows:
- Creative advertisements, influence customers or consumers to buy the product.
- Advertisings involve information regarding the product and so increases product awareness.
- Advertising on social media platforms, TVs, radio and newspapers, promotes company branding.
Hence, the correct options are:
- influencing consumer tastes
- increasing product awareness
- promoting company branding
Answer: 11.2 moles of 
are produced when 5.60 mol of ethane is burned in an excess of oxygen.
Explanation:
The combustion of ethane is represented using balanced chemical equation:
As oxygen is preset in excess, ethane acts as the limiting reagent as it limits the formation of product.
According to stoichiometry :
2 moles of propane produces 4 moles of carbon dioxide
Thus 5.60 moles of propane will produce=
moles of carbon dioxide
Thus 11.2 moles of are produced when 5.60 mol of ethane is burned in an excess of oxygen.
This is false, because copper is an element.
To be solution, it has to have at least 2 different substances. Solid solution of metals is often called alloy.
The pH indicators to be used are Phenolphthalein, Red cabbage, Bromthymol blue and Congo red.
<h3>What are pH indicators?</h3>
Indicators are substances which change color as the pH of a medium changes.
The common indicators and their pH range is as follows:
- Phenolphthalein - pH range of 8.3 and 10.5
- Red cabbage - pH 2 to 10
- Bromthymol blue - 6.0 to 7.6
- Congo red - 3.0 to 5.2
Therefore, the indicators to be used are Phenolphthalein, Red cabbage, Bromthymol blue and Congo red.
Learn more about pH indicators at: brainly.com/question/13779537
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Please, you have to apply the formula below:<span>Q=c∗m∗Δt</span>where Q is the energy lost, c is the specific heat of water, m is the mass of water involved, so m=3.75 *10^-1 Kg c=4,184 J/(Kg*°C) delta t=37.5 °C
Taking density of water as 1000kg/m3. Mass of water would be 0.375kg. So, heat lost would be<span>H=mCDeltaT</span>H=0.375*4184*37.5 = 58837.5J
