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3 years ago

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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3 years ago

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Lina20 [59]

Answer:

Please see attachment

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Please see attachment

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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 integrated rate law for a first-order reaction 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

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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 add heat to boil water

ΔS > 0 (positive), because changing from a liquid to a gas increases the disorder

ΔG = 0, because the liquid-vapour equilibrium process is at equilibrium 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