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daser333 [38]
2 years ago
5

Determine the maximum amount of NaNO3 that was produced during the experiment. Explain how you determine this amount.

Chemistry
2 answers:
Natali5045456 [20]2 years ago
8 0

Answer:

9 moles

Explanation:

see image linked!

tatiyna2 years ago
7 0

Answer:

9 moles

Explanation:

3 moles from NaNO3, 3 moles from Al(NO3)3

3 × 3 = 9 moles

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How is an atomic mass unit defined?
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A unit of mass used to express atomic and molecular weights, equal to one-twelfth of the mass of an atom of carbon-12. It is equal to approximately 1.66 x 10-27<span> kg.</span>
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Do you know how to do this? if you have 4 grams of carbon-14, how much carbon 14 remains after 11,460 years? the half life is 5,
Delvig [45]
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3 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
3 years ago
As a general rule, how many specific substrates can bind to an enzyme?
san4es73 [151]

Answer:

Enzyme Active Site and Substrate Specificity

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5 0
3 years ago
Read 2 more answers
B. If the sand you ran across has a specific-heat capacity of 835 J/(kgºc),
Murrr4er [49]

Answer: 16700 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed = ?

m = mass of sand = 2 kg

c = heat capacity = 835J/kg^0C

Initial temperature  = T_i = 40^0C

Final temperature= T_f  = 50^0C

Change in temperature ,\Delta T=T_f-T_i=(50-40)^0C=10^0C

Putting in the values, we get:

Q=2kg\times 835J/kg^0C\times 10^0C

Q=16700J

16700 J of energy must be added to a 2-kilogram pile of it to increase its temperature from 40°C to 50°C

4 0
3 years ago
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