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3 years ago

Find the acceleration if a car goes from 40 to 50 mph in 5 seconds

Chemistry

1 answer:

Anastaziya [24]3 years ago

3 0

Answer:

62.5 mph/seconds

Explanation:

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PLEASE HELP!!!! 60 points

lina2011 [118]

Answer:

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3 years ago

Read 2 more answers

In 24.02g of carbon there are _blank_atoms of carbon?

andriy [413]

The answer is two atoms of carbon.

8 0

4 years ago

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.350 M , [B] = 0.650 M , and [C] = 0.300 M . Th

uysha [10]

<u>Answer:</u> The value of equilibrium constant, $K_c$ for the given reaction is 12.85.

<u>Explanation:</u>

For the given chemical equation:

$A+2B\rightleftharpoons C$

At t = 0 0.350M 0.650M 0.300M

At $t=t_{eq}$ (0.350 - x) (0.650 - 2x) (0.300 + x)

We are given:

Equilibrium concentration of A = 0.220 M

Forming an equation for concentration of A at equilibrium:

$0.350-x=0.220\\x=0.130$

Thus, the concentration of B at equilibrium becomes = $0.650-(2\times 0.130)=0.390M$

Equilibrium concentration of C = 0.430 M

The expression of $K_c$ for the given chemical equation is:

$K_c=\frac{[C]}{[A][B]^2}$

Putting values in above equation:

$K_c=\frac{0.430}{0.220\times (0.390)^2}\\\\K_c=12.85$

Hence, the value of equilibrium constant, $K_c$ for the given reaction is 12.85.

8 0

3 years ago

How much Cu-61 (half-life about 3 hours) would remain from a 2 mg sample after 9 hours?

vivado [14]

Answer: 0.25mg

Explanation:

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3 years ago

Calculate the Gibbs energy of freezing (DeltaGfreezing) in units of J/mol when supercooled water freezes at -3degreeC at constan

frutty [35]

Explanation:

It is known that the change in Gibb's free energy varies with temperature as follows.

$\Delta G(T) = \Delta H(T) - T \Delta S(T)$

= $\Delta H(T_{f}) - \Delta C_{p,m} (T - T_{f}) - T[\Delta S(T_{f}) - \Delta C_{p,m} ln (\frac{T}{T_{f}})]$

$\Delta H(T_{f}) = -\Delta_{fus} H(T_{f})$ (assumption)

= $\Delta H(T_{f}) - \frac{T}{T_{f}} \Delta H(T_{f}) - \Delta C_{p, m}(T - T_{f} - T ln \frac{T}{T_{f}})$

= $(\frac{T}{T_{f}} - 1) \Delta_{fus} H(T_{f}) - \Delta C_{p,m}(T - T_{f} - Tln (\frac{T}{T_{f}}))$

As, T = $-3^{o}C$ = (-3 + 273) = 270 K, $T_{f} = 0^{o}C = 0 + 273 K = 273 K$ .

Therefore, calculate the change in Gibb's free energy as follows.

$\Delta G(T) = (\frac{T}{T_{f}} - 1) \Delta_{fus} H(T_{f}) - \Delta C_{p,m}(T - T_{f} - Tln (\frac{T}{T_{f}}))$

= $(\frac{270 K}{273 K} - 1)(6000 J/mol K) - (75.3 - 38) J/mol K (270 K - 273 K - 270 K ln \frac{270 K}{273 K})$

= -65.93 J/mol K + 0.62 J/mol K

= -65.31 J/mol K

Thus, we can conclude that Gibbs energy of freezing for the given reaction is -65.31 J/mol K.

6 0

4 years ago