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dlinn [17]
3 years ago
9

When discussing acids and bases any substance that accepts a patron by definition is considered an

Chemistry
1 answer:
vredina [299]3 years ago
8 0

Brønsted-Lowry acid.

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A sample of 0.600 mol of a metal m reacts completely with excess fluorine to form 46.8 g of mf2. how many moles of f are in the
oksian1 [2.3K]
The balanced chemical reaction is expressed as:

M + F2 = MF2

To determine the moles of the element fluorine present in the product, we need to determine the moles of the product formed from the reaction and relate this value to the ratio of the elements in MF2. We do as follows:

moles MF2 produced = 0.600 mol M ( 1 mol MF2 / 1 mol M ) = 0.600 mol MF2
molar mass MF2 = 46.8 g MF2 / 0.6 mol MF2 = 78 g/mol
moles MF2 = 46.8 g ( 1 mol / 78 g ) = 0.6 mol
moles F = 0.6 mol MF2 ( 2 mol F / 1 mol MF2 ) = 1.2 moles F
6 0
3 years ago
Use dimensional analysis to convert 14.5mi/hr to km/s
Liula [17]

Answer:

0.006 48 km/s

Explanation:

1. Convert miles to kilometres

14.5 mi × (1.609 km/1 mi) = 23.33 km

2. Convert hours to seconds

1 h × (60 min/1h) × (60 s/1 min) = 3600 s

3. Divide the distance by the time

14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s

5 0
4 years ago
Why does sulfur have a higher melting point than phosphorus
tiny-mole [99]
Because phosphorus is more flammable than sulfur!

Have a nice day! :)

5 0
4 years ago
1. Put the following steps in order, please number 1‐6.
Tresset [83]

Answer:

1.hypothesis

2.ask a question

3.experiment

4.collect observations

5.analyze data

6.draw conclusion

Explanation:

8 0
3 years ago
Upon landing, the 90.7 kg carbon fiber brakes of an airliner heat up 312∘C, producing heat. As the brakes start to cool back to
Crank

Given,

Mass of carbon fiber brakes = 90.7 kg = 90.7 x 1000 = 90700 g  

Mass of the rubber tires = 123 kg = 123 x 1000 = 123000

Specific heat of carbon fiber = 1.400Jg∘C

As all the heat is transferred from the carbon fiber brakes to the rubber tires, the heat lost by the carbon fiber brakes is equal to the heat gained by the rubber tires.

So the temperature change of carbon fiber brakes, delta T = 172∘C

And the temperature change of rubber tires, delta T = 172∘C

The formula used for heat is,  

Q = m s delta T

where Q is heat, m is mass, s is specific heat, and delta T is the temperature change

We know,  

Heat lost by the carbon fiber brakes = Heat gained by the rubber tires

90700 g x 1.400Jg∘C x 172∘C = 123000 g x s x 172∘C

or,  90700 g x 1.400Jg∘C = 123000 g x s

or, s =  (90700 g x 1.400Jg∘C) / 123000 g

or, s = 1.03236

Therefore, the specific heat of rubber tires is 1.03236 Jg∘C

8 0
3 years ago
Read 2 more answers
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