Ask question

Ask question

All categories

4 years ago

5

The half life of a new isotope found on Mars is 2 years. It takes 10 years to get the sample back to Earth. If the astronauts st

arted with 128g of
this new isotope how much is left when they get back to Earth?

1 answer:

yawa3891 [41]4 years ago

3 0

Answer:

3.99 g

Explanation:

The following data were obtained from the question:

Half life (t½) = 2 years.

Original amount (N₀) = 128 g

Time (t) = 10 years

Amount remaining (N) =..?

Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:

Half life (t½) = 2 years.

Decay constant (K) =.?

K= 0.693/t½

K = 0.693/2

K = 0.3465 year¯¹.

Finally, we shall determine the amount remaining after 10 years i.e the amount remaining when they arrive on Earth. This can be obtained as follow:

Original amount (N₀) = 128 g

Time (t) = 10 years

Decay constant (K) = 0.3465 year¯¹.

Amount remaining (N) =..?

Log (N₀/N) = kt/2.3

Log (128/N) = (0.3465 × 10)/2.3

Log (128/N) = 1.5065

Take the antilog of 1.5065

128/N = Antilog (1.5065)

128/N = 32.1

Cross multiply

128 = 32.1 × N

Divide both side by 32.1

N = 128/32.1

N = 3.99 g

Therefore, the amount remaining is 3.99 g

You might be interested in

The bond angles in SCl2 are expected to be Multiple Choice a little more than 109.5°. 109.5°. 120°. a little less than 109.5°. 1

nalin [4]

Answer:

a little less than 109.5°

Explanation:

SCl2 has four regions of electron density around the central atom of the molecule. This implies that it has a tetrahedral electron domain geometry with an expected bond angle of 109.5° according to valence shell electron pair repulsion theory.

However, there are two lone pair of electrons on the central atom of the molecule which decreases the bond angle a little less than 109.5° owing to repulsion between electron pairs.

4 0

3 years ago

How do I do these problems

horsena [70]

you calculate all the letters together and it would be Zn2

8 0

3 years ago

SHOW YOUR WORK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

WARRIOR [948]

Answer:

1.375%

Explanation:

Percent Error = measured value - accepted value/ accepted value x 100

Measured Value: 15.78

Accepted Value: 16.00

Work:

$\frac{15.78-16.00}{16.00}$ * 100

$\frac{-.22}{16}$ * 100 = -1.375

You cannot have a negative percentage. Therefore the answer is 1.375%

4 0

3 years ago

rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is

Sunny_sXe [5.5K]

Complete Question

The rate of a certain reaction is given by the following rate law:

$rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$ ?

What is the reaction order in $I_2$ ?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in $H_2$ is n = 1

The reaction order in $I_2$ is m = 1

The overall reaction order z = 2

When the hydrogen is double the the initial rate is $rate_n = 4.0*10^{-4} M/s$

The rate constant is $k = 35.23 \ M^{-1} s^{-1}$

Explanation:

From the question we are told that

The rate law is $rate = k [H_2][I_2]$

The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be n and for $I_2$ be m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is n=m=1

The overall reaction order is

$z = n +m$

$z =1 +1$

$z =2$

At $rate = 2.0 *10^{4} M /s$

$2.0*10^{4} = k [H_2] [I_2] ---(1)$

= > $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

$rate = k [2H_2] [I_2] ----(2)$

=> $k = \frac{rate_n }{ [2H_2] [I_2]}$

So equating the two k

$\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

=> $rate_n = 4.0*10^{-4} M/s$

So when

$rate_x = 52.0 M/s$

$[H_2] = 1.8 M$

$[I_2] = 0.82 \ M$

We have

$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

$k = 35.23 M^{-2} s^{-1}$

3 0

3 years ago

A zinc block with a mass of 230 g is given 1320 J of energy. What is the change in

Mariana [72]

Answer:

14.7°C

Explanation:

Q = m·ΔT·c

ΔT = $\frac{Q}{m*c}$

ΔT = $\frac{1320 J}{230 g* 0.39 J/gC}$

= 1320 J / ((230 g) * (.39 J/g°C)

ΔT = 14.7 °C

4 0

3 years ago