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IceJOKER [234]
3 years ago
8

What is the definition of an alternative acid?

Chemistry
1 answer:
Fantom [35]3 years ago
6 0

alternative acids are more cost-effective and allow for flavor expression through nontraditional methods and ingredients, making for increased versatility.

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Can someone help me with this? (WILL GIVE BRAINLIEST)
iragen [17]

the answer is in the picture, btw the molar mass for the first one is wrong, it should be 77.98, and the final product is 2.32

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3 years ago
what is different between an ionic bond and and a covalent bond? a. an ionic bond shares electrons, and a covalent bond is an at
BaLLatris [955]
I would pick either b) or D) but you just have to remember that not all ionicbonds will bond together, because some bonds are only negattive bonds, 
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3 years ago
The orbital radius of Venus is 0.72 AU. What is this distance in kilometers? (One AU is about 150 million kilometers.)
Evgen [1.6K]
<span>The answer is D) 108 million kilometers. To solve this problem, you must perform a simple unit conversion calculation. 1 AU = 150,000,000 km is the conversion factor. Take the radius of Venus, .72 AU, and multiply it by 150,000,000 km/1 AU. You flip the conversion factor so that the units of the original distance in the numerator cancel the units in the denominator of the conversion factor. completing the calculation gives you 108,000,000 km</span>
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3 years ago
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Hitman42 [59]

Answer:

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6 0
3 years ago
One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from t
larisa [96]
The balanced chemical equation is written as:

<span>CsF(s) + XeF6(s) ------> CsXeF7(s)

We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride.  We calculate as follows:

11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
6 0
3 years ago
Read 2 more answers
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