What percentage of pure alcohol is in a shot of 100 proof liquor?

1 answer:

7 0

<span>First of all, an ounce of hard liquor (gin, vodka, rum, whiskey, or scotch) contains 64 calories for 80 proof varieties, and 80 calories for those that are100 proof. The proof refers to the percentage of alcohol in the liquor (80 proof is 40 percent alcohol; 100 proof is 50 percent alcohol)</span>

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How do you calculate the mass of a product when the amounts of more than one reactant are given?

chubhunter [2.5K]

you add the masses of the reactants, because of conservation of mass. if there are two or more products they will ask you to find the mass of only one product or the sum of the mass of all products

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3 years ago

An enzyme is: a. A biological reactant b. A biological product c. A biological catalyst d. a biological co-factor

vovangra [49]

It is <span>a substance produced by a living organism that acts as a catalyst to bring about a specific biochemical reaction.
So I think C matches the discription </span>

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2 years ago

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The impacts of global warming are being felt and will be felt increasingly in your lifetime. What impacts are likely to be seen

IceJOKER [234]

Answer:

Hotter climate, more greenhouse gases, and less water

I hope this helps :)

Explanation:

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3 years ago

f 37.4 grams of water decompose at 297 Kelvin and 1.30 atmospheres, how many liters of oxygen gas can be produced?

Allisa [31]

2H2O=2H2+O2
37.4g H2O(1 mol/18.02)=2.07547 mol H2O
PV=nRT
(1.30)(V)=(2.07547)(.0821)(297)
Vwater=38.92898L
38.92898L (1 mol O2/2 mol H2O)=19.46449L O2 gas

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3 years ago

Copper(II) fluoride contains 37.42% F by mass. Calculate the mass of fluorine (in g) in 55.5 g of copper(II) fluoride.

Mrac [35]

Answer:

There are 20.8 g of fluorine in 55.5 g of copper (II) fluoride

Explanation:

x % by mass of a species in a specimen means there are x g of the species in total 100 g of a specimen

37.42 % F by mass means 100 g of copper (II) fluoride contains 37.42 g of F.

So, 100 g of copper (II) fluoride contains 37.42 g of F

55.5 g of copper (II) fluoride contains $\frac{37.42\times 55.5}{100}$ g of F or 20.8 g of F

Hence there are 20.8 g of fluorine in 55.5 g of copper (II) fluoride.

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3 years ago