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3 years ago

10

Earth is approximately 1.5x10^8 km from the sun. The speed of light is 3.0x10^8. How many minutes does it take for light to reac

h earth from the sun?

1 answer:

Masteriza [31]3 years ago

3 0

1.5 x 10 $^{8}$ mm = 1500 x 10 $^{8}$ m

1500 x 10 $^{8}$ / ( 3.0 x 10 $^{8}$ ) = 500 s

500/60 = 8.3

So approximately it takes 8.5 minutes

Hope that helps, Good luck! (:<span />

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A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant K_f = 5.32 de

kirill115 [55]

Answer: $-15.4^00C$

Explanation:

$T^0_f-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = boiling point of solution = ?

$T^o_f$ = boiling point of solvent (X) = $-10.1^oC$

$k_f$ = freezing point constant = $5.32^oC/kgmol$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte like urea)

= mass of solute (urea) = 29.82 g

= mass of solvent (X) = 500.0 g

$M_2$ = molar mass of solute (urea) = 60 g/mol

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

$(-10.1-(T_f))^oC=1\times (5.32^oC/m)\times \frac{(29.82g)\times 1000}{60\times (500.0g)}$

$T_f=-15.4^0C$

Therefore, the freezing point of solution is $-15.4^0C$

7 0

3 years ago

Let the simulator run for 30 seconds. What happens to the size of the ice?

Shkiper50 [21]

10 is your answer hope it helps you.......... Xd

<h2>Answer : 10 </h2>

7 0

3 years ago

How many chlorine atoms are there in one molecule of PCI3

Darya [45]

Answer:

Chemical Formula:

For example- for every one mole of the compound PCl3 P C l 3 , it contains 3 moles of Chlorine atom and 1 mole of Phosphorus atom.

Explanation:

• Phosphorus trichloride (PCl3) contains three chlorine atoms and one phosphorus atoms.

8 0

2 years ago

Which of the following is true about a system at equilibrium? a The concentration(s) of the reactant(s) is equal to the concentr

anyanavicka [17]

Answer:

c The concentration(s) of reactant(s) is constant over time.

Step-by-step explanation:

When the reaction A ⇌ B reaches equilibrium, the concentrations of reactants and products are constant over time.

a is <em>wrong</em>, because the concentrations of reactants and products are usually quite different.

b is <em>wrong</em>, because both product and reactant molecules are being formed at equilibrium.

d is <em>wrong</em>. The rates of the forward and reverse reactions are equal, but they are not zero.

6 0

3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

3 years ago