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elixir [45]
3 years ago
10

Which quantity represents 0.500 Mole at STP

Chemistry
1 answer:
KIM [24]3 years ago
4 0
0.500 moles is roughly .5*6.022*10^23=3.011*10^23 atoms. This is independent of STP.
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if you could place a piece of solid silver into a container of liquid silver, would it float or sink? Explain your answer
olga55 [171]
Liquid silver is less dense than solid silver, so the solid silver would sink.
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In a similar experiment, unlabeled 4-butanolide was allowed to stand in an acidic solution in which the water had been labeled w
Lina20 [59]

Answer:

Please see attachment

Explanation:

Please see attachment

6 0
3 years ago
How long will it take for a 750 mg sample of radium with a half life of 15 days to decay to exactly 68mg?
weqwewe [10]

Answer:

52 da  

Step-by-step explanation:

Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.

The i<em>ntegrated rate law for a first-order reaction </em>is  

ln([A₀]/[A] ) = kt

Data:

[A]₀ = 750 mg

 [A] =    68 mg

t_ ½ =   15 da

Step 1. Calculate the value of the rate constant.

 t_½ = ln2/k     Multiply each side by k

kt_½ = ln2         Divide each side by t_½

      k = ln2/t_½

         = ln2/15

         = 0.0462 da⁻¹

Step 2. Calculate the time

ln(750/68) = 0.0462t

         ln11.0 = 0.0462t

            2.40 = 0.0462t     Divide each side by 0.0462

                   t = 52 da

8 0
3 years ago
The rate constant for a certain reaction is measured at two different temperatures:
Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

E_a= activation energy = ?

R= gas constant = 8.314 J/kmol

T_1 = temperature = 280.0^0C=(273+280)=553K

T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

3 0
3 years ago
Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp
Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

         Equilibrium when T = 100 °C

      Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must <em>add heat</em> to boil water

ΔS > 0 (positive), because changing from a liquid to a gas i<em>ncreases the disorder </em>

ΔG = 0, because the liquid-vapour equilibrium process is at <em>equilibrium</em> at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.


If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.


If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0
3 years ago
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