Answer:
The picture is most likely showing fishing/ fishery. other industries that rely on accurate weather forecast are shrimping, aquaculture, and other ocean industries
Explanation:
Answer:
2578.99 years
Explanation:
Given that:
100 g of the wood is emitting 1120 β-particles per minute
Also,
1 g of the wood is emitting 11.20 β-particles per minute
Given, Decay rate = 15.3 % per minute per gram
So,
Concentration left can be calculated as:-
C left =
Where,
is the concentration at time t
is the initial concentration
Also, Half life of carbon-14 = 5730 years
Where, k is rate constant
So,
The rate constant, k = 0.000120968 year⁻¹
Time =?
Using integrated rate law for first order kinetics as:
So,
<u>t = 2578.99 years</u>
There are quite a bunch of typo errors in the question; here is the correct question below:
A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass = 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T = 300 K? Take an average diameter of d = 3.1 × 10⁻¹⁰ m for the molecules in the air.
Answer:
9.57 × 10⁻⁷ atm
Explanation:
The mean free path ( λ ) can be illustrated by the equation:
λ =
---------- (1)
N/V =
------------- (2)
From the above relation, we can deduce that;
P=
-------------(3)
let I= λ
From the above equations;
d= diameter of the atom
= avogadro's constant
P= pressure
R= universal rate constant which is given by 0.08206 L atm mol⁻¹ k⁻¹
T= temperature
From the question, we are given that the mean free path of the residual air molecules ( d = 3.1 × 10⁻¹⁰ m) is equal to 10cm = 0.1m
Therefore, we can determine the pressure using equation (3)
i.e
P=
= ![\frac{(8.314J{K^-^1)(300K)}}{\sqrt{2}(3.14)(3.1*10^{-10}m)^2(6.022*10^{23}mol^{-1})(0.1m) }](https://tex.z-dn.net/?f=%5Cfrac%7B%288.314J%7BK%5E-%5E1%29%28300K%29%7D%7D%7B%5Csqrt%7B2%7D%283.14%29%283.1%2A10%5E%7B-10%7Dm%29%5E2%286.022%2A10%5E%7B23%7Dmol%5E%7B-1%7D%29%280.1m%29%20%7D)
=97.06 × 10⁻³ Pa × ![\frac{1atm}{1.01325*10^5Pa}](https://tex.z-dn.net/?f=%5Cfrac%7B1atm%7D%7B1.01325%2A10%5E5Pa%7D)
=9.57 × 10⁻⁷ atm
Therefore, the maximum residual gas pressure in the calculated space is; 9.57 × 10⁻⁷ atm
Answer:
Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions.
Explanation:
hope this help :)
The given question is incomplete. The complete question is as follows.
Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is the oxidation of ammonia. In this reaction, gaseous ammonia reacts with dioxygen gas to produce nitrogen monoxide gas and water.
Suppose a chemical engineer studying a new catalyst for the oxidation of ammonia reaction finds that 645. liters per second of dioxygen are consumed when the reaction is run at 195.oC and 0.88 atm. Calculate the rate at which nitrogen monoxide is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.
Explanation:
Chemical equation for the oxidation of ammonia is as follows.
![4NH_{3}(g) + 5O_{2}(g) \rightleftharpoons 4NO(g) + 6H_{2}O(l)](https://tex.z-dn.net/?f=4NH_%7B3%7D%28g%29%20%2B%205O_%7B2%7D%28g%29%20%5Crightleftharpoons%204NO%28g%29%20%2B%206H_%7B2%7DO%28l%29)
Then volume of
per second consumed is as follows.
V = ![\frac{645 L}{1 sec}](https://tex.z-dn.net/?f=%5Cfrac%7B645%20L%7D%7B1%20sec%7D)
As this reaction is taking place at a temperature of
(468.15 K) and pressure 0.88 atm. Hence, moles of consumption of
are calculated as follows.
n = ![\frac{PV}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7BPV%7D%7BRT%7D)
= ![\frac{0.88 atm \times 645 L/sec}{0.08206 L atm/mol K \times 468.15 K}](https://tex.z-dn.net/?f=%5Cfrac%7B0.88%20atm%20%5Ctimes%20645%20L%2Fsec%7D%7B0.08206%20L%20atm%2Fmol%20K%20%5Ctimes%20468.15%20K%7D)
= 14.77 mol
/sec
When 5 moles of
produces 4 moles of NO then the amount of NO produced from 14.77 mol
![14.77 mol O_{2} /sec \times \frac{4 mol NO}{5 mol O_{2}} \times \frac{30.01 g NO}{mol NO}](https://tex.z-dn.net/?f=14.77%20mol%20O_%7B2%7D%20%2Fsec%20%5Ctimes%20%5Cfrac%7B4%20mol%20NO%7D%7B5%20mol%20O_%7B2%7D%7D%20%5Ctimes%20%5Cfrac%7B30.01%20g%20NO%7D%7Bmol%20NO%7D)
= 354.60 g/s
Therefore, NO formed per second is as follows.
![354.60 g/s \times \frac{kg}{10^{3} g}](https://tex.z-dn.net/?f=354.60%20g%2Fs%20%5Ctimes%20%5Cfrac%7Bkg%7D%7B10%5E%7B3%7D%20g%7D)
= 0.35 kg/s
Thus, we can conclude that the rate at which nitrogen monoxide is being produced is 0.35 kg/s.