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Sunny_sXe [5.5K]
3 years ago
7

Which is least likely to occur after an experiment i conducted to test a hypothesis?

Chemistry
1 answer:
Alenkinab [10]3 years ago
8 0

Answer:

The hypothesis becomes a theory if the results support it is answer.

Explanation:

I hope it's helpful!

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What amount (moles) is represented by each of these samples?
Zinaida [17]
<h3>Answer:</h3>

                  a)  Moles of Caffeine  =  1.0 × 10⁻⁴ mol

                  b) Moles of Ethanol   =  4.5 × 10⁻³ mol

<h3>Solution:</h3>

Data Given:

                  Mass of Caffeine  =  20 mg  =  0.02 g

                  M.Mass of Caffeine  =  194.19 g.mol⁻¹

                  Molecules of Ethanol  =  2.72 × 10²¹

Calculate Moles of Caffeine as,

                               Moles  =  Mass ÷ M.Mass

Putting values,

                               Moles  =  0.02 g ÷ 194.19 g.mol⁻¹

                                Moles  =  1.0 × 10⁻⁴ mol

Calculate Moles of Ethanol as,

                                                         As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Molecules ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  2.72 × 10²¹ Molecules ÷ 6.022 × 10²³

                          Number of Moles  =  4.5 × 10⁻³ Moles

5 0
3 years ago
Really strong acids and bases are found at the​
makkiz [27]

Ends....................?

4 0
3 years ago
Energy in a chemical reaction can be identified as which of the following?
daser333 [38]
I believe it’s A..but I’m not quite sure.
6 0
2 years ago
How many grams of potassium (K) contain 5.11 x 10^22 atoms of potassium?
Darina [25.2K]

The atomic mass of K is 39

from Avogadro's law

39g of K contains 6.02x10^23 atoms

therefore if

39=6.02x19^23

X=5.11×10^22

making X the subject of the formula

X= (5.11×10^22×39)÷6.02×10^23

X= 33g

7 0
2 years ago
When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If t
satela [25.4K]

Answer:

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

Explanation:

Heat lost by solution ad calorimeter = Q

Heat capacity of solution ad calorimeter = C = 1071 J/°C

Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C

Q=C\times Delta T

Q=1071 J/^oC\times (-3.44^oC)=-3,684.24 J

Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J

Moles of sodium nitrate = \frac{15.3 g}{85 g/mol}=0.18 mol

When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.

Then heat absorbed by 1 mole of sodium nitrate :

\frac{3,684.24 J}{0.18}=20,468 J=20.468 kJ

1 J = 0.001 kJ

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

8 0
3 years ago
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