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3 years ago

13

Please help! I put all my points into this question, brainliest, thanks, rattings, etc.

1 answer:

kkurt [141]3 years ago

8 0

A higher concentration of reactants leads to more collisions per unit time and leads to an increasing reaction rate. Changing the pressure of gaseous reactants is, in effect, changing their concentration. The increased number of collisions caused by a higher pressure generally increases the reaction rate.

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Uranium-235 decays to form lead-207. The half-life of uranium-235 is 704,000,000 years. What is the approximate age of an igneou

Strike441 [17]

Answer:

The approximate age of the igneous rock is $7.042\times 10^{6}$ years.

Explanation:

Chemically speaking, the Uranium-235 decays to form Lead-207 throughout time. All isotopes decay exponentially by means of the following model:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (1)

In addition, we can determine the time constant of the former isotope in terms of its half life, that is:

$\tau = \frac{t_{1/2}}{\ln 2}$ (2)

Where:

$m_{o}$ - Initial mass of Uranium-235, in atoms.

$m(t)$ - Current mass of Uranium-235, in atoms.

$\tau$ - TIme constant, in years.

$t_{1/2}$ - Half-life of Uranium-235, in years.

Please remind that a 1 : 1 ratio of Uranium-235 to Lead-207 atoms means that current mass of Uranium-235 is a half of its initial mass. If we know that $m(t) = 0.5\cdot m_{o}$ and $t_{1/2} = 7.04\times 10^{8}\,yr$ , then the approximate age of the igneous rock is:

$\tau = \frac{7.04\times 10^{8}\,yr}{\ln 2}$

$\tau \approx 1.016\times 10^{9}\,yr$

$t = -\tau \cdot \ln \left(\frac{m(t)}{m_{o}} \right)$

$t = -(1.016\times 10^{9}\,yr)\cdot \ln \frac{1}{2}$

$t \approx 7.042\times 10^{6}\,yr$

The approximate age of the igneous rock is $7.042\times 10^{6}$ years.

5 0

3 years ago

How to solve the equation for the value of X

rodikova [14]

Answer:

first, to get rid of the division, multiply both sides by 2. you will get 7x=98. next, divide both sides by 7 to isolate x. x=14

7 0

3 years ago

How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?

natali 33 [55]

I think the answer is D

4 0

3 years ago

A graduated cylinder contains 18.0ml of water. what is the new water level after 35.6g of silver metal with a density of 10.5 g/

AlekseyPX

Answer:

The answer to your question is

1.- Volume = 3.4 ml

2.- Volume = 0.61 ml

3.- Mass = 2872.8 pounds

Explanation:

Problem 1

Volume = 18 ml

mass = 35.6 g

density = 10.5 g/ml

Process

1.- Calculate the volume of silver

Formula

$density = \frac{mass}{volume}$

solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{35.6}{10.5}$

<u>volume = 3.4 ml</u>

2.- Problem 2

Total volume = ?

Volume = 18 + 3.4

Volume = 21.4 ml

Data

mass = 8.3 g

density = 13.6 g(ml

volume = ?

Formula

$density = \frac{mass}{volume}$

Solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{8.3}{13.6}$

Result

<u>volume = 0.61 ml</u>

3.- Problem 3

Data

volume = 345 gal

density = 1 g/ml

mass = ?

Formula

$density = \frac{mass}{volume}$

Solve for mass

mass = density x volume

Covert gal to ml

1 gal --------------- 3785 ml

345 gal ------------- x

x = (345 x 3785) / 1

x = 1305825 ml

Substitution

mass = 1 x 1305825

mass = 1305825 g

Convert g to pounds

1 g ------------------- 0.0022 pounds

1305825 g ---------------- x

x = (1305825 x 0.0022)

<u> x = 2872.8 pounds</u>

5 0

3 years ago

Which of the following best describes scientists?

ratelena [41]

Answer:

B

Explanation:

Man with variety backgrounds

8 0

1 year ago

Read 2 more answers