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Luden [163]
3 years ago
14

A(n) 816 ml sample of hydrogen was collected over water at 21.0◦c on a day when the barometric pressure was 755 torr. what volum

e would the dry hydrogen occupy under standard conditions? the vapor pressure of water at 21.0◦c is 19.0 torr. answer in units of ml.
Chemistry
1 answer:
hammer [34]3 years ago
6 0
The correct answer is A

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Vanyuwa [196]

Answer:

it identifies reducing sugars (monosaccharide's and some disaccharides), which have free ketone or aldehyde functional group

Explanation:

It turns from turquoise to yellow or orange when it reacts with reducing sugars.

3 0
2 years ago
A 2.36-gram sample of NaHCO3 was completely decomposed in an experiment.
enyata [817]

Answer:

Explanation:

a) The mass of the reactants is 2.36 grams, and the mass of the products is 1.57 grams plus the mass of the carbonic acid. Thus, using the law of conservation of mass, we get the mass of the carbonic acid is 2.36 - 1.57 =  0.79 grams.

b) The gram-formula mass of sodium bicarbonate is 84.006 g/mol, meaning that 2.36/84.006 = 0.028 moles were consumed. Thus, this means that in theory, 0.014 moles of carbonic acid should have been produced, which would have a mass of (0.014)(62.024)=0.868 grams. Thus, the percentage yield is (0.79)/(0.868) * 100 = 91%

8 0
2 years ago
Which of the following is the correct chemical formula for cs and br? csbr cs2br csbr2
nalin [4]
The correct chemical formulae is CsBr
4 0
2 years ago
Read 2 more answers
The activation energy of a reaction is 56.9 kj/mol and the frequency factor is 1.5×1011/s. Part a calculate the rate constant of
Vanyuwa [196]

We will use Arrehenius equation

lnK = lnA  -( Ea / RT)

R = gas constant = 8.314 J / mol K

T = temperature = 25 C = 298 K

A = frequency factor

ln A = ln (1.5×10 ^11) = 25.73

Ea = activation energy = 56.9 kj/mol = 56900 J / mol

lnK = 25.73 - (56900 / 8.314 X 298) = 2.76

Taking antilog

K = 15.8



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vichka [17]

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3 years ago
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