the half-life of a certain radioactive element is 1250 years. what percent of the atom remains after 7500 years?

If a swimming pool contains 2,850 kiloliters (kL) of water, how many gallons (gal) of water does it contain? (1 gal = 3.785 L)

siniylev [52]

we are given

a swimming pool contains 2,850 kiloliters (kL) of water

2,850 kiloliters (kL)=2850000L

we know that

$1gal=3.785L$

$3.785L=1gal$

Firstly , we will find for 1L

$1L=\frac{1}{3.785}gal$

now, we can multiply both sides by 2850000

$2850000*1L=2850000*\frac{1}{3.785}gal$

$2850000L=2850000*\frac{1}{3.785}gal$

$2850000L=752972.25892gal$

so,

a swimming pool contains 752972.25892gal of water...........Answer

6 0

2 years ago

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Carbon (C): 1sH2sI2pJ H = I = J =

ratelena [41]

Answer:H=2 I=2 J=2

Explanation:

4 0

2 years ago

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Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM

7nadin3 [17]

Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: The value of Kc remains the same.

As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, The reaction must: A. run in the forward direction to reestablish equilibrium.

By decreasing the volume, and so that Kc remains constant, being:

$Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2} ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume} } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume$

where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so, the number of moles of Cl₂ should decrease.The number of moles of Cl2 will B. decrease.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, the value of Qc C. is less than Kc.

3 0

2 years ago

Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d

PilotLPTM [1.2K]

Answer:

k = 23045 N/m

Explanation:

To find the spring constant, you take into account the maximum elastic potential energy that the spring can support. The kinetic energy of the car must be, at least, equal to elastic potential energy of the spring when it is compressed to its limit. Then, you have:

$K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2$ (1)

M: mass of the car = 1050 kg

k: spring constant = ?

v: velocity of the car = 8 km/h

x: maximum compression of the spring = 1.5 cm = 0.015m

You solve the equation (1) for k. But first you convert the velocity v to m/s:

$v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}$

$k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}$

The spring constant is 23045 N/m

3 0

2 years ago

How many moles of AgNO3 must react to form 0.854 mol Ag?

kipiarov [429]

The answer:
The equation of its dissolution in water is: AgNO3 → Ag + (aq) + NO3- (aq)

and AgNO3 → Ag + (aq) + NO3- (aq)
1 mol 1mol 1mol
? -------- 0.854mo
so for finding the value, it is sufficients to complute 1 x 0.854 mol =0.854 mol
so, 0.854 mol is required for the reaction to form 0.854 mol of Ag

5 0

2 years ago

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