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Sonja [21]
3 years ago
10

In 2005, a Nevada silver mine exported 200 tons of silver. In 2008, as the silver supply became exhausted, it only exported 25 t

ons of silver. What is the rate of change of the mine’s silver output? A. -175/3. B. -1/50. C. 175/3. D. 75
Chemistry
2 answers:
lianna [129]3 years ago
8 0
The rate of change in the amount of silver supplied is equal to the ratio of the difference in the amount supplied to the number of years elapsed from 2005 to 2008. That is,
               rate of change = (25 - 200) / (2008 - 2005) = -173/3
The answer is letter A. 
Anit [1.1K]3 years ago
5 0

Answer:

The correct answer is option A.

Explanation:

Amount of silver exported in 2005 = 200 tons

Amount of silver exported in 2008 = 25 tons

Rate of change of silver output:

\frac{\text{Silver exported in 2008}-\text{Silver exported in 2005-}}{\text{time elapsed}}=\frac{25-200}{2008-2005}=\frac{-175}{3}

Hence, the correct answer is option A.

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Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife​
yuradex [85]

Answer:

\boxed{\text{2408 min}}

Explanation:

The integrated rate law for radioactive decay is

\ln\dfrac{N_{0}}{N_{t}} = kt

1. Calculate the decay constant

\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}

8 0
3 years ago
In each case, calculate the appropriate ratio to show that the information given is consistent with the law of multiple proporti
Alchen [17]

Answer:

(a) 3:2; (b) 2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

(a) Ammonia and hydrazine.

In ammonia, the mass ratio of H:N is r₁ = 0.2158/1

In hydrazine, the mass ratio of H:N is r₂ = 0.1439/1

The ratio of the ratios is:

r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.2158}{0.1439} = \dfrac{1.500 }{1} = \dfrac{2.999}{2} \approx \mathbf{\dfrac{3}{2}}\\\\\text{The relative amounts of H in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{3}{2}}}

(b) Nitrogen oxides

In nitrogen monoxide, the mass ratio of O:N is r₁ = 1.142/1

In dinitrogen monoxide, the mass ratio of O:N is r₂ = 0.571/1

The ratio of the ratios is:

r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 1.142}{0.571} = \dfrac{2.000 }{1} \approx \mathbf{\dfrac{2}{1}}\\\\\text{The relative amounts of O in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}

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A + B + C → D + E
brilliants [131]
I believe it is B... because all balance equation are supposed to follow the law of conservation of mass

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How many grams are in 3.78 x 10^22 molecules of SO2?
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Answer:

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